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Online Dilution Calculator: Chemistry Solvers

Solve dilution problems with wolfram|alpha, more than just an online dilution calculator.

Wolfram|Alpha is a great tool for calculating the molarity, molality, mass fraction and amount fraction concentration of solutions. It also generates solution properties and preparation recipes.

Dilution calculator example images

Learn more about:

  • Chemical solutions

Tips for entering queries

Enter your queries using plain English. Here are some examples illustrating how to ask for a dilution.

  • dilute Vi = 25.0 mL, Ci = 2.19M, Vf=72.8 mL
  • dilute 25 ml of 2.19 M to 72.8 ml
  • dilute 0.75 L of 0.250 M KOH with 2.00 L of water
  • molarity 2.004 mg NaAlO2 in 500 ml water
  • concentrated hydrogen chloride solution
  • prepare 2.5 mM ammonium molybdate
  • View more examples

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What are dilutions?

Dilution is an important experimental technique for creating solutions of a desired concentration..

A solution is a mixture of components that is homogenous at the molecular level. The solution components are two or more pure substances that were mixed to form the solution. For a binary solution containing two components, the minor component is called the solute and the other component is the solvent . The solute concentration describes the amount of solute dispersed in a given quantity of the solvent. Dilution is a process whereby the concentration of the solute in a stock solution is reduced by the addition of more solvent. Diluting a solution resulting from a previous dilution by adding the same amount of solvent is known as a serial dilution. Dilution calculations allow one to determine how much solvent to add or what the final concentration is after the addition of solvent.

DILUTION_IMAGE

How Wolfram|Alpha calculates dilutions

Wolfram|Alpha computes dilutions using the dilution equation, , or the ideal dilution equation, , by calling FormulaData . The dilution equation is applicable for all dilutions since the non-additivity of solution volumes is accounted for. Solution volumes are not additive due to the intermolecular forces between the solute and solvent entities. If the intermolecular forces between the solute and the solvent are the same as the intermolecular forces in the pure solvent, the solution is said to be ideal and the ideal dilution equation can be used. In practice, the ideal dilution formula is only applicable for solutions of very low concentration or situations where errors in concentration do not affect experimental results.

moles solute before dilution = moles solute after dilution
M = moles of solute / volume of solution
M 1 V 1 = M 2 V 2
(1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) x = 100. mL
(2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x) x = 454.5454545 mL (oops, my fingers got stuck typing.) (Bad attempt at humor, really bad!) x = 454.5 mL
(0.750 mol/L) (100.0 mL) = (1.00 mol/L) (x) x = 75.0 mL
(0.100 mol/L) (100.0 mL) = (0.250 mol/L) (x) Please go ahead and solve for x.
(1.50 mol/L) (x) = (0.800 mol/L) (2.00 L) x = 1.067 L Divide the liters by 1000 to get mL, the answer is 1070 mL (notice that it is rounded off to three sig figs)
1) calculate total moles 2) calculate total volume 3) divide moles by volume to get molarity
M 1 V 1 + M 2 V 2 = M 3 V 3
MV = grams / molar mass (x) (0.0220 L) = 3.10 g / 165.998 g/mol x = 0.84886 M
M 2 V 1 = M 2 V 2 (4.05 mol/L) (51.0 mL) = (0.84886 mol/L) (y) y = 243 mL (to three sig figs)
AuNO 3 ---> Au + + NO 3 ¯ Al(NO 3 ) 3 ---> Al 3+ + 3NO 3 ¯ (10.00 mL) x (0.200 mmol/mL AuNO 3 ) x (1 mol NO 3 ¯ / 1 mol AuNO 3 ) = 2.00 mmol NO 3 ¯ from AuNO 3 (15.00 mL) x (0.180 mmol/mL Al(NO 3 ) 3 ) x (3 mol NO 3 ¯ / 1 mol Al(NO 3 ) 3 ) = 8.10 mmol NO 3 ¯ from Al(NO 3 ) 3 Note the alternate definition of molarity, that being mmol/mL.
(2.00 mmol + 8.10 mmol) / (10.00 mL + 15.00 mL) = 0.404 mmol/mL = 0.404 mol/L NO 3 ¯
C 1 V 1 = C 2 V 2 (48.3 mL) (11.6 mg/dL) = (3.19 mg/dL) (V 2 ) V 2 = (48.3 mL x 11.6 mg/dL) / (3.19 mg/dL) = 175.6 mL total
(175.6 mL total) − (48.3 mL initially) = 127.3 mL to be added
(A) One solution is 0.1487 M and has a density of 1.018 g/mL (B) The other solution is 10.00%(w/w) and has a density of 1.038 g/mL.
(A) This solution has a concentration of 0.1487 mole/L and 1.00 L of it weighs 1018 g. (0.1487 mol/L) (1.00 L) = 0.1487 mol (moles of solute) (0.1487 mol) (342.2948 g/mol) = 50.9 g (mass of solute) 1018 g − 50.9 g = 967.1 g (mass of solvent) 967.1 g / 18.015 g/mol = 53.683 mol (moles of solvent)
(1038 g) (0.1000) = 103.8 g (mass of solute) 103.8 g / 342.2948 g/mol = 0.30325 mol (moles of solute) 1038 − 103.8 = 934.2 g (mass of solvent) 934.2 g / 18.015 g/mol = 51.8568 mol (moles of solvent)
total moles sucrose ---> 0.1487 + 0.30325 = 0.45195 mol total moles ---> 53.683 + 51.8568 + 0.1487 + 0.30325 = 105.99175 mol mole fraction ---> 0.45195 mol / 105.99175 mol = 0.004264
(0.004264) (100) = 0.4264%
(a) Calculate the concentration (in molarity) of the final solution. (b) Calculate the mass of KMnO 4 needed to directly prepare the final solution.
MV = grams / molar mass (x) (0.5000 L) = 0.9597 g / 158.032 g/mol x = 0.01214564 M
M 1 V 1 = M 2 V 2 (0.01214564 M) (2.000 mL) = (y) (1000. mL) <>y = 0.0000242913 M
M 1 V 1 = M 2 V 2 (0.0000242913 M) (10.00 mL) = (z) (250.0 mL) z = 0.000000971652 M To four sig figs and in scientific notation: 9.716 x 10¯ 7 M
250.0 mL is the volume of the final molarity MV = grams / molar mass (9.71652 x 10¯ 7 mol/L) (0.2500 L) = x / 158.032 g/mol x = 0.00003839 g (to four sig figs)

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Dilution Example Problems

Dilution

In most laboratory settings, a stock solution is created when a compound is used over and over. This stock solution will have a high concentration. If lower concentrations are needed, a dilution is performed.

A dilution is a process where the concentration of a solution is lowered by adding solvent to the solution without adding more solute. These dilution example problems show how to perform the calculations needed to make a diluted solution.

The key idea behind a dilution is the number of moles of solute in the solutions does not change as the solvent is added.

moles of solute prior to dilution = moles solute after dilution

The concentration of a solution can be expressed in molarity (M).

Dilution math 1

Solve for moles and get:

Since moles of solute prior to dilution = moles solute after dilution,

M i V i = M D V D where: M i = initial concentration V i = initial volume M D = diluted concentration V D = diluted volume

Example Problem 1:

Problem: What volume of 5 M NaOH is needed to create a 100 mL solution of 1 M NaOH?

Solution: Use the formula M i V i = M D V D .

M i = 5 M V i = initial volume M D = 1 M V D = 100 mL

Dilution Math step

Answer: 20 mL of 5 M NaOH is needed to create a 100 mL solution of 1 M NaOH. Water is added to the 20 mL solution until there is 100 mL.

Example Problem 2:

Problem: If you have 300 mL of 1.5 M NaCl, how many mL of 0.25 M NaCl can you make?

Solution:  Use the formula M i V i = M D V D .

M i = 1.5 M V i = 300 mL M D = 0.25 M V D = final volume

Dilution algebra step

Answer: You can make 1800 mL of 0.25 M NaCl solution from 300 mL of 1.5 M NaCl solution.

Related Posts

Calculating Concentrations with Units and Dilutions

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Calculating the concentration of a chemical  solution  is a basic skill all students of chemistry must develop early in their studies. What is concentration? Concentration refers to the amount of solute that is dissolved in a solvent . We normally think of a solute as a solid that is added to a solvent (e.g., adding table salt to water), but the solute could easily exist in another phase. For example, if we add a small amount of ethanol to water, then the ethanol is the solute, and the water is the solvent. If we add a smaller amount of water to a larger amount of ethanol, then the water could be the solute.

How to Calculate Units of Concentration

Once you have identified the solute and solvent in a solution, you are ready to determine its concentration . Concentration may be expressed several different ways, using percent composition by mass , volume percent , mole fraction , molarity , molality , or normality .

Percent Composition by Mass (%)

This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100. Example: Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt. Solution: 20 g NaCl / 100 g solution x 100 = 20% NaCl solution

Volume Percent (% v/v)

Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as: v/v % = [(volume of solute)/(volume of solution)] x 100% Note that volume percent is relative to the volume of the solution, not the volume of solvent . For example, wine is about 12% v/v ethanol. This means there is 12 ml ethanol for every 100 ml of wine. It is important to realize liquid and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution. As another example, 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).

Mole Fraction (X)

 This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1. Example: What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92) Solution: 90 g water = 90 g x 1 mol / 18 g = 5 mol water 92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glycerol total mol = 5 + 1 = 6 mol x water = 5 mol / 6 mol = 0.833 x glycerol = 1 mol / 6 mol = 0.167 It's a good idea to check your math by making sure the mole fractions add up to 1: x water + x glycerol = .833 + 0.167 = 1.000

Molarity (M)

Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!). Example: What is the molarity of a solution made when water is added to 11 g CaCl 2 to make 100 mL of solution? (The molecular weight of CaCl 2 = 110) Solution: 11 g CaCl 2 / (110 g CaCl 2 / mol CaCl 2 ) = 0.10 mol CaCl 2 100 mL x 1 L / 1000 mL = 0.10 L molarity = 0.10 mol / 0.10 L molarity = 1.0 M

Molality (m)

Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water. Example: What is the molality of a solution of 10 g NaOH in 500 g water? (Molecular weight of NaOH is 40) Solution: 10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH 500 g water x 1 kg / 1000 g = 0.50 kg water molality = 0.25 mol / 0.50 kg molality = 0.05 M / kg molality = 0.50 m

Normality (N)

Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capacity of a given molecule. Normality is the only concentration unit that is reaction dependent. Example: 1 M sulfuric acid (H 2 SO 4 ) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H + ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

  • Grams per Liter (g/L) This is a simple method of preparing a solution based on grams of solute per liter of solution.
  • Formality (F) A formal solution is expressed regarding formula weight units per liter of solution.
  • Parts per Million (ppm) and Parts per Billion (ppb) Used for extremely dilute solutions, these units express the ratio of parts of solute per either 1 million parts of the solution or 1 billion parts of a solution. Example: A sample of water is found to contain 2 ppm lead. This means that for every million parts, two of them are lead. So, in a one gram sample of water, two-millionths of a gram would be lead. For aqueous solutions, the density of water is assumed to be 1.00 g/ml for these units of concentration.

How to Calculate Dilutions

You dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:

M i V i = M f V f

where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.

Example: How many milliliters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

Solution: 5.5 M x V 1 = 1.2 M x 0.3 L V 1 = 1.2 M x 0.3 L / 5.5 M V 1 = 0.065 L V 1 = 65 mL

So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume

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PLANETCALC Online calculators

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Dilution calculator and problems solver

This online calculator can calculate the molar concentration (molarity) of a solute or volume of a solution before or after the dilution.

These online calculators can help with dilution problems. Generally, in dilution problems, you either dilute a solution or mix two solutions with different concentrations. So, the first calculator below can solve dilution problems, and the second calculator below can solve mix problems. Theory and formulas can be found below the calculators.

Dilute a solution problems solver

This calculator can solve the following types of problems:

  • Find the final molarity of a solution
  • Find the final volume of a solution
  • Find the initial molarity of a solution
  • Find the initial volume of a solution

To see examples for each type of problem, change the problem type in the calculator below.

PLANETCALC, Dilution calculator and problems solver

Mix solutions problems solver

  • Find the molarity of the final solution
  • Find the volume of the final solution
  • Find the molarity of one of the starting solutions

PLANETCALC, Mixed solutions calculator and problems solver

Mixed solutions calculator and problems solver

First starting solution, second starting solution, final solution, dilution and molarity.

Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent, like adding more water to a solution. To dilute a solution means to add more solvent without the addition of more solute. The resulting solution is thoroughly mixed to ensure that all parts of the solution are identical.

Solution concentration is described with molarity (or molar concentration), which is the number of moles of solute per liter of solution, measured in mol/liter, denoted as M, and calculated as follows:

M=\frac{\text{mol solute}}{\text{L of solution}}

which can be rearranged like this

M \cdot \text{L of solution}=\text{mol solute}

which gives us the following proportion

M_s \cdot V_s=M_f \cdot V_f

This equation is used in the first calculator. For example, to find final molarity, you use

M_f =\frac{M_s \cdot V_s}{V_f}

The same logic can be applied to the mix problems. Of course, we assume that the problems are about mixing the same solute, and there is no volume contraction upon mixing.

Thus, the equation look like follows:

M_{s_1} \cdot V_{s_1} + M_{s_2} \cdot V_{s_2}=M_f \cdot V_f

From these equations, it is easy to find the final values or unknown concentration of one of the starting solutions.

Sometimes dilution problem gives amounts of solute in grams. In this case, you need to know the solute's chemical formula and then convert the mass of the solute to the number of moles or find molarity. You can use Molar mass of the substance and Molarity calculator for this.

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Biology LibreTexts

4: Dilution Worksheet and Problems

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  • Understand how to quantify bacterial cells.
  • Learn how to solve a dilution problem.

It is a common practice to determine microbial counts for both liquid and solid specimens--suspensions of E. coli in nutrient broth all the way to soil samples and hamburger meat. Most specimens have high enough numbers of microorganisms that the specimen has to be serially diluted to quantitate effectively. The following is a step-by-step procedure to working dilution problems, and includes some practice problems at the end.

The purpose can be determination of bacterial, fungal, or viral counts (indirectly). This protocol is specific for bacterial counts (colony-forming units, CFUs), but can be modified for fungi (CFUs) and viruses (plaque-forming units, PFUs for viral counts). A set of serial dilutions is made, a sample of each is placed into a liquefied agar medium, and the medium poured into a petri dish. The agar solidifies, with the bacterial cells locked inside of the agar. Colonies grow within the agar, as well as on top of the agar and below the agar (between the agar and the lower dish). The procedure described above produces a set of pour plates from many dilutions, but spread plates (sample spread on top of solidified agar) can be used also. The agar plate allows accurate counting of the microorganisms, resulting from the equal distribution across the agar plate. This cannot be done with a fluid solution since 1) one cannot identify purity of the specimen, and 2) there is no way to enumerate the cells in a liquid.

dilution1.png

SOLVING DILUTION PROBLEMS

colony count on agar plate

THE STANDARD FORMULA = ___________________________________________________________

total dilution of tube (used to make plate for colony count) X amount plated

To work the problem, you need 3 values---a colony count from the pour or spread plates, a dilution factor for the dilution tube from which the countable agar plate comes, and the amount of the dilution that was plated on the agar plate.

STEP 1:Determine the appropriate plate for counting

Look at all plates and find the one with 30-300 colonies (or plaques), preferably. Greater than 300 and less than 30 is a high degree of error. Air contaminants can contribute significantly to a really low count and a high count can be confounded by error in counting too many small colonies. Use the total dilution for the tube from where the plate count was obtained. If duplicate plates (with same amount plated) have been made from one dilution, average the counts together.

STEP 2:Determine the total dilution for the dilution tubes

Dilution = amount of specimen transferred divided by the [amount of specimen transferred + amount already in tube].

Determine the dilution factor for each tube in the dilution series. Multiply the individual dilution of the tube X previous total dilution. To calculate this dilution series:

dilution3.png

amount of sample

Dilution factor for each tube in a set = _______________________________________

amount of sample + amount of diluent in tube

But after the first tube, each tube is a dilution of the previous dilution tube.

Total dilution factor = previous dilution of tube X dilution of next container

Example: FOR THE ABOVE DILUTION SERIES

1 ml added to 9ml = 1/10 for 1st tube

1ml added to 9ml = 1/10 for 2nd tube

previous dilution of 1/10 (1st tube) X 1/10 (2nd tube) = total dilution of 1/100 (=10 -2 =1/10 2 )

STEP 3: Determine the amount plated

The amount plated is the amount of dilution used to make the particular pour plate or spread plate.

There is nothing to calculate here: the value will be stated in the procedure, or it will be given in the problem.

dilution4.png

Solving the above problem

  • The countable plate is the one with 71 colonies.
  • The total dilution of 3 rd tube from which above pour plate was made = 1/10 X 1/10 X 1/10 = 1/10 3
  • The amount used to make that pour plate = 1ml

71 colonies

__________ = 71 X 10 3 = 7.1 X 10 4 (scientific notation) OR 71,000/ml

Note: Rules for Scientific Notation

dilution5.png

For a number to be in correct scientific notation, the following conditions must be true:

  • The coefficient must be greater than or equal to 1 and less than 10.
  • The base must be 10.
  • The exponent must show the number of decimal places that the decimal needs to be moved to change the number to standard notation. A negative exponent means that the decimal is moved to the left when changing to standard notation.

1. Determine the number of bacteria per ml. of water specimen.

dilution6.png

2. Determine the number of bacterial cells per ml. in the original culture.

dilution7.png

3. Determine the number of bacterial cells per gram of meat.

dilution8.png

Contributors

Jackie Reynolds, Professor of Biology ( Richland College )

Concentrations of Solutions Quiz

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This online quiz is intended to give you extra practice in determining the concentrations of solutions or performing dilution calculations.

Select your preferences below and click 'Start' to give it a try!

Solution Dilution Calculator

Dilute solution of known molarity.

The solution dilution calculator tool calculates the volume of stock concentrate to add to achieve a specified volume and concentration. The calculator uses the formula M 1 V 1 = M 2 V 2 where "1" represents the concentrated conditions (i.e., stock solution molarity and volume) and "2" represents the diluted conditions (i.e., desired volume and molarity). To prepare a solution of specific molarity based on mass, please use the Mass Molarity Calculator. To dilute a solution of concentrated acid or base of known w/w% strength, please use the Acid & Base Molarity Calculator.

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Solution Dilution Calculator

Table of contents

The solution dilution calculator will calculate for you how to dilute a stock solution of known concentration to obtain an arbitrary volume of a diluted solution. Read this article to learn how to use this tool correctly and what units to choose. And if you're interested in chemistry, give our half-life calculator a look!

How to calculate dilution

The main objective of this concentration calculator is determining how to dilute a stock solution. Imagine you have a concentrated solution of hydrochloric acid. You can use this calculator to determine how much of it you need if you want to obtain 200 mL of a diluted solution with a concentration of 20 mM.

You can easily calculate this value, using the following dilution formula:

  • m 1 m_1 m 1 ​ — The concentration of stock solution;
  • m 2 m_2 m 2 ​ — The concentration of diluted solution;
  • V 1 V_1 V 1 ​ — The volume of the stock solution; and
  • V 2 V_2 V 2 ​ — The volume of diluted solution.

Note that this equation is not equivalent to the proportion formula .

Units of concentration

You probably know that the units of volume are either cubed units of length (for example, cubic meters, cubic millimeters, etc.) or liters. We will just remind you that 1 liter is equal to 1 cubic decimeter.

What about the units of concentration, then? You can either use molar or mass concentration. Our solution dilution calculator uses molar concentration, but we will teach you how to recalculate the units of mass concentration, too.

  • Molar concentration c c c is the amount of substance in moles in a given volume of substance. It is expressed in the unit "molar" (symbol: M \mathrm{M} M ), where 1   M = 1   m o l / l i t e r 1\ \mathrm{M} = 1\ \mathrm{mol/liter} 1   M = 1   mol/liter .
  • Mass concentration ρ \rho ρ is the amount of substance in grams in a given volume of substance. It is expressed in gram per liter.

If you want to find the mass concentration of your solution, you need to multiply the molar concentration by the molar mass of the substance M (expressed in kg/mol):

You can also find it using our molarity calculator .

💡 Did you know, reconstitution produces a proper solution using a dry and wet ingredient! Check out the reconstitution calculator to know more.

How to use the solution dilution calculator

  • Determine the concentration of the stock solution. Let's say it is equal to 1 1 1 mol per liter, or 1   M 1\ \mathrm M 1   M .
  • Decide on the final volume of the solution you want to obtain. Let's say you want 0.5 0.5 0.5 liters of it.
  • Decide on the concentration of the obtained solution. Let's say you want it to be equal to 20   m M 20\ \mathrm{mM} 20   mM .
  • Input all this data into the dilution equation:
  • You can also use the solution dilution calculator to obtain any other value. Simply type the remaining three into the corresponding boxes.

🙋 Did you know that for solutions, the mass percent represents the ratio of solute to the total solution?

Quantum physicist's take on boiling the perfect egg. Includes times for quarter and half-boiled eggs.

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Concentration (initial)

Volume (initial)

Concentration (final)

Volume (final)

Calculators for preparing physiological and biochemical solutions

  • C 1 is the concentration of the stock solution.
  • V 1 is the volume to be removed (i.e., aliquoted) from the concentrated stock solution.
  • C 2 is the final concentration of the diluted solution.
  • V 2 is the final volume of the diluted solution. This is the volume that results after V 1 from the stock solution has been diluted with diluent to achieve a total diluted volume of V 2 .
  • An alternative and commonly-used notation for this equation is M 1 V 1 = M 2 V 2 , where M is used in place of C .

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Chemistry LibreTexts

4.5: Molarity and Dilutions

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  Learning Objectives

  • Describe the fundamental properties of solutions
  • Calculate solution concentrations using molarity
  • Perform dilution calculations using the dilution equation

In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (Figure \(\PageIndex{1}\)). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

A picture is shown of sugar being poured from a spoon into a cup.

We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration . Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved . Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution .

A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity ( M ) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.2} \]

Example \(\PageIndex{1}\): Calculating Molar Concentrations

A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?

Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:

\[\begin{align*} M &=\dfrac{mol\: solute}{L\: solution} \\[4pt] &=\dfrac{0.133\:mol}{355\:mL\times \dfrac{1\:L}{1000\:mL}} \\[4pt] &= 0.375\:M \label{3.4.1} \end{align*} \]

Exercise \(\PageIndex{1}\)

A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?

Example \(\PageIndex{2}\): Deriving Moles and Volumes from Molar Concentrations

How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example \(\PageIndex{1}\)?

In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 3.4.2, 0.375 M :

\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.3} \]

\[ \begin{align*} \mathrm{mol\: solute} &= \mathrm{ M\times L\: solution} \label{3.4.4} \\[4pt] \mathrm{mol\: solute} &= \mathrm{0.375\:\dfrac{mol\: sugar}{L}\times \left(10\:mL\times \dfrac{1\:L}{1000\:mL}\right)} &= \mathrm{0.004\:mol\: sugar} \label{3.4.5} \end{align*} \]

Exercise \(\PageIndex{2}\)

What volume (mL) of the sweetened tea described in Example \(\PageIndex{1}\) contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?

Example \(\PageIndex{3}\): Calculating Molar Concentrations from the Mass of Solute

Distilled white vinegar (Figure \(\PageIndex{2}\)) is a solution of acetic acid, \(CH_3CO_2H\), in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:

\[\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=\dfrac{25.2\: g\: \ce{CH3CO2H}\times \dfrac{1\:mol\: \ce{CH3CO2H}}{60.052\: g\: \ce{CH3CO2H}}}{0.500\: L\: solution}=0.839\: \mathit M} \label{3.4.6} \]

\[M=\mathrm{\dfrac{0.839\:mol\: solute}{1.00\:L\: solution}} \nonumber \]

Nov 29, 2019, 5:24 PM

\[\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=0.839\:\mathit M} \label{3.4.7} \]

Exercise \(\PageIndex{3}\)

Calculate the molarity of 6.52 g of \(CoCl_2\) (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

Example \(\PageIndex{4}\): Determining the Mass of Solute in a Given Volume of Solution

How many grams of NaCl are contained in 0.250 L of a 5.30- M solution?

The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example \(\PageIndex{3}\):

\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.9} \]

\[\mathrm{mol\: solute= \mathit M\times L\: solution} \label{3.4.10} \]

\[\mathrm{mol\: solute=5.30\:\dfrac{mol\: NaCl}{L}\times 0.250\:L=1.325\:mol\: NaCl} \label{3.4.11} \]

Finally, this molar amount is used to derive the mass of NaCl:

\[\mathrm{1.325\: mol\: NaCl\times\dfrac{58.44\:g\: NaCl}{mol\: NaCl}=77.4\:g\: NaCl} \label{3.4.12} \]

Exercise \(\PageIndex{4}\)

How many grams of \(CaCl_2\) (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?

5.55 g \(CaCl_2\)

When performing calculations stepwise, as in Example \(\PageIndex{3}\), it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example \(\PageIndex{4}\), the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (Example \(\PageIndex{5}\)). This eliminates intermediate steps so that only the final result is rounded.

Example \(\PageIndex{5}\): Determining the Volume of Solution

In Example \(\PageIndex{3}\), we found the typical concentration of vinegar to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid?

First, use the molar mass to calculate moles of acetic acid from the given mass:

\[\mathrm{g\: solute\times\dfrac{mol\: solute}{g\: solute}=mol\: solute} \label{3.4.13} \]

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

\[\mathrm{mol\: solute\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.14} \]

Combining these two steps into one yields:

\[\mathrm{g\: solute\times \dfrac{mol\: solute}{g\: solute}\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.15} \]

\[\mathrm{75.6\:g\:\ce{CH3CO2H}\left(\dfrac{mol\:\ce{CH3CO2H}}{60.05\:g}\right)\left(\dfrac{L\: solution}{0.839\:mol\:\ce{CH3CO2H}}\right)=1.50\:L\: solution} \label{3.4.16} \]

Exercise \(\PageIndex{5}\):

What volume of a 1.50-M KBr solution contains 66.0 g KBr?

Dilution of Solutions

Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure \(\PageIndex{2}\)).

Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution , we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure \(\PageIndex{3}\)).

A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters:

\[n=ML \nonumber \]

Expressions like these may be written for a solution before and after it is diluted:

\[n_1=M_1L_1 \nonumber \]

\[n_2=M_2L_2 \nonumber \]

where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, n 1 = n 2 . Thus, these two equations may be set equal to one another:

\[M_1L_1=M_2L_2 \nonumber \]

This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:

\[C_1V_1=C_2V_2 \nonumber \]

where \(C\) and \(V\) are concentration and volume, respectively.

Example \(\PageIndex{6}\): Determining the Concentration of a Diluted Solution

If 0.850 L of a 5.00- M solution of copper nitrate, Cu(NO 3 ) 2 , is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?

We are given the volume and concentration of a stock solution, V 1 and C 1 , and the volume of the resultant diluted solution, V 2 . We need to find the concentration of the diluted solution, C 2 . We thus rearrange the dilution equation in order to isolate C 2 :

\[C_2=\dfrac{C_1V_1}{V_2} \nonumber \]

Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M . We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:

\[C_2=\mathrm{\dfrac{0.850\:L\times 5.00\:\dfrac{mol}{L}}{1.80\: L}}=2.36\:M \nonumber \]

This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M ).

Exercise \(\PageIndex{6}\)

What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL?

0.102 M \(CH_3OH\)

Example \(\PageIndex{7}\): Volume of a Diluted Solution

What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?

We are given the volume and concentration of a stock solution, V 1 and C 1 , and the concentration of the resultant diluted solution, C 2 . We need to find the volume of the diluted solution, V 2 . We thus rearrange the dilution equation in order to isolate V 2 :

\[V_2=\dfrac{C_1V_1}{C_2} \nonumber \]

Since the diluted concentration (0.12 M ) is slightly more than one-fourth the original concentration (0.45 M ), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:

\[V_2=\dfrac{(0.45\:M)(0.011\: \ce L)}{(0.12\:M)} \nonumber \]

\[V_2=\mathrm{0.041\:L} \nonumber \]

The volume of the 0.12- M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate.

Exercise \(\PageIndex{7}\)

A laboratory experiment calls for 0.125 M \(HNO_3\). What volume of 0.125 M \(HNO_3\) can be prepared from 0.250 L of 1.88 M \(HNO_3\)?

Example \(\PageIndex{8}\): Volume of a Concentrated Solution Needed for Dilution

What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?

We are given the concentration of a stock solution, C 1 , and the volume and concentration of the resultant diluted solution, V 2 and C 2 . We need to find the volume of the stock solution, V 1 . We thus rearrange the dilution equation in order to isolate V 1 :

\[V_1=\dfrac{C_2V_2}{C_1} \nonumber \]

Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M ), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:

\[V_1=\dfrac{(0.100\:M)(5.00\:\ce L)}{1.59\:M} \nonumber \]

\[V_1=0.314\:\ce L \nonumber \]

Thus, we would need 0.314 L of the 1.59- M solution to prepare the desired solution. This result is consistent with our rough estimate.

Exercise \(\PageIndex{8}\)

What volume of a 0.575-M solution of glucose, C 6 H 12 O 6 , can be prepared from 50.00 mL of a 3.00-M glucose solution?

0.261 

Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.

Key Equations

  • \(M=\mathrm{\dfrac{mol\: solute}{L\: solution}}\)
  • C 1 V 1 = C 2 V 2

IMAGES

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VIDEO

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COMMENTS

  1. ChemTeam: Dilution Problems #1-10

    Calculate the molar concentration of the final solution. 1) Calculate moles HCl in 0.445 M solution: (0.445 mol/L) (2.00 L) = 0.890 moles 2) Set up expression for moles of HCl in second solution: (x) (3.88 L) = moles HCl in unknown solution 3) Calculate moles of HCl in final solution: (0.974 mol/L) (5.88 L) = 5.73 moles

  2. 7.19: Concentrations: Dilution

    Because the final volume of the solution, V2, is larger than the calculated initial volume, V1, the value of the final answer is reasonable. Exercise 7.19.1 7.19. 1. Calculate the concentration that results upon diluting 250 milliliters of a 3.9 M solution to a final volume of 1.55 liters. Answer.

  3. 6.1.1: Practice Problems- Solution Concentration

    PROBLEM 6.1.1.6 6.1.1. 6. Calculate the molarity of each of the following solutions: (a) 0.195 g of cholesterol, C 27 H 46 O, in 0.100 L of serum, the average concentration of cholesterol in human serum. (b) 4.25 g of NH 3 in 0.500 L of solution, the concentration of NH 3 in household ammonia.

  4. 11.4: Dilutions and Concentrations

    Concentration is the removal of solvent, which increases the concentration of the solute in the solution. (Do not confuse the two uses of the word concentration here!) In both dilution and concentration, the amount of solute stays the same. This gives us a way to calculate what the new solution volume must be for the desired concentration of ...

  5. Dilution Calculator: Wolfram|Alpha Chemistry Solvers

    Dilution is an important experimental technique for creating solutions of a desired concentration. A solution is a mixture of components that is homogenous at the molecular level. The solution components are two or more pure substances that were mixed to form the solution.

  6. Dilution (video)

    We can relate the concentrations and volumes before and after a dilution using the following equation: M₁V₁ = M₂V₂ where M₁ and V₁ represent the molarity and volume of the initial concentrated solution and M₂ and V₂ represent the molarity and volume of the final diluted solution. Created by Sal Khan. Questions Tips & Thanks

  7. Dilution Problems, Chemistry, Molarity & Concentration Examples

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  8. Dilution Problems

    This is a chemistry tutorial that covers dilution problems, including examples of how to calculate the new concentration of a diluted solution, and how to calculate the volume of a...

  9. ChemTeam: Dilution

    Example #7: Calculate the final concentration if 2.00 L of 3.00 M NaCl, 4.00 L of 1.50 M NaCl and 4.00 L of water are mixed. Assume there is no volume contraction upon mixing. The solution to this problem is almost exactly the same as 10a. The only "problem child" appears to be the 4.00 L of water.

  10. Dilution Example Problems

    This stock solution will have a high concentration. If lower concentrations are needed, a dilution is performed. A dilution is a process where the concentration of a solution is lowered by adding solvent to the solution without adding more solute. These dilution example problems show how to perform the calculations needed to make a diluted ...

  11. Calculating Concentrations with Units and Dilutions

    How to Calculate Units of Concentration Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using percent composition by mass, volume percent, mole fraction, molarity, molality, or normality . Percent Composition by Mass (%)

  12. 13.7: Solution Dilution

    The dilution is by a factor of 32 to go from 16M 16 M to 0.5M 0.5 M. Exercise 13.7.1 13.7. 1. A 0.885 M solution of KBr with an initial volume of 76.5 mL has more water added until its concentration is 0.500 M.

  13. Online calculator: Dilution calculator and problems solver

    This online calculator can calculate the molar concentration (molarity) of a solute or volume of a solution before or after the dilution. These online calculators can help with dilution problems. Generally, in dilution problems, you either dilute a solution or mix two solutions with different concentrations.

  14. PDF Dilution and Concentration

    = + Pharmaceutical preparation before dilution Pharmaceutical preparation after dilution (same amount of active drug although the volume has doubled) Problems such as these sometimes seem complicated and difficult. Solving some of these calculations requires a series of steps.

  15. 4: Dilution Worksheet and Problems

    STEP 2:Determine the total dilution for the dilution tubes. Dilution = amount of specimen transferred divided by the [amount of specimen transferred + amount already in tube]. Determine the dilution factor for each tube in the dilution series. Multiply the individual dilution of the tube X previous total dilution.

  16. Concentrations of Solutions Quiz : ChemQuiz.net

    This online quiz is intended to give you extra practice in determining the concentrations of solutions or performing dilution calculations. Select your preferences below and click 'Start' to give it a try! This quiz helps you practice calculating the molarity (concentration) of a chemical solution with a variety of options, including dilutions.

  17. Solution Dilution Calculator

    This solution dilution calculator tool calculates the volume of stock concentrate to add to achieve a specified volume and concentration using the formula M1V1 = M2V2.

  18. Solution Dilution Calculator

    You can use this calculator to determine how much of it you need if you want to obtain 200 mL of a diluted solution with a concentration of 20 mM. You can easily calculate this value, using the following dilution formula: m_1\cdot V_1 = m_2\cdot V_2 m1 ⋅ V 1 = m2 ⋅ V 2. where: m 1. m_1 m1.

  19. 4.13: Dilutions and Concentrations

    The concentration of the solution has decreased. In going from 25.0 mL to 72.8 mL, 72.8 − 25.0 = 47.8 mL of solvent must be added. Exercise 4.13.1 4.13. 1. A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution? Answer.

  20. How solve dilution and concentration calculation problems

    This video illustrates how to solve dilution and concentration calculation problems. Share this video: https://youtu.be/2Vx9oYeHQqcSubscribe to my channel to...

  21. Dilution Calculator

    C2 is the final concentration of the diluted solution. V2 is the final volume of the diluted solution. This is the volume that results after V1 from the stock solution has been diluted with diluent to achieve a total diluted volume of V2. An alternative and commonly-used notation for this equation is M1V1 = M2V2, where M is used in place of C.

  22. 13.7: Solution Dilution

    The dilution is by a factor of 32 to go from 16M 16 M to 0.5M 0.5 M. Exercise 13.7.1 13.7. 1. A 0.885 M solution of KBr with an initial volume of 76.5 mL has more water added until its concentration is 0.500 M.

  23. Serial Dilution

    A serial dilution is a series of sequential dilutions used to reduce a dense culture of cells to a more usable concentration. The easiest method is to make a series of 1 in 10 dilutions.

  24. 4.5: Molarity and Dilutions

    Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: M = molsolute Lsolution. Example 4.5.1: Calculating Molar Concentrations. A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar).