Some Basic Concepts of Chemistry · Lesson 10 of 11

Expressing Concentration in Solutions

“Choose the correct concentration measure and use it to prepare, dilute, and react solutions.”

Learning Objectives

• Distinguish solution mass, solvent mass, and solution volume. • Calculate mass percentage, mole fraction, molarity, and molality. • Explain dilution as conservation of solute amount. • Convert concentration measures using suitable mass and density data. • Use molarity with a balanced equation in a solution reaction.

How much solute is present?

Many reactions occur in solutions, where reactants are distributed through a solvent. Knowing only the solution’s total mass or volume does not reveal the amount of reacting substance. Concentration measures describe the relative amount of a component, but different measures use different denominators.

Definition

Solute and solvent

The solute is the component being dissolved or the component whose amount is being specified. The solvent is the dissolving medium. Together they form the solution.

For a simple solution made by adding a solid to water, the solution mass is solute mass plus water mass. Its final volume need not equal the sum of the separately measured component volumes. Always read whether a question says solvent, solution, or final volume; these words determine the calculation.

Mass percentage and mole fraction

Mass percentage compares solute mass with total solution mass. It is useful when quantities have been weighed. Mole fraction instead compares one component’s amount with the total amount of all components. These are different proportions because equal masses of different substances do not generally contain equal numbers of molecules.

Mass percentage of a soluteLaTeX

Mass % = \frac{m _{solute}}{m _{solution}} \times 100

m_{solution} = m_{solute} + m_{solvent}

for a two-component solution. Both masses must use the same unit.

Example — A weighed solution

Problem
Find the solute mass percentage when $2.00 g$ of A is dissolved in $18.0 g$ water.

1.Total solution mass is $2.00 + 18.0 = 20.0 g$ at the reported precision.
2.Mass percentage is $(2.00/20.0) 100 = 10.0%$ .
3.Dividing by 18.0 would compare solute with solvent, not with the whole solution. That would not be mass percentage as defined here.

Mole fractionLaTeX

x_{A} = \frac{n _{A}}{n _{A} + n _{B}} x_{B} = \frac{n _{B}}{n _{A} + n _{B}} x_{A} + x_{B} = 1

n_{A}

and

n_{B}

are amounts in moles. For more components, include every component in the denominator. Mole fraction is dimensionless.

Example — A molecular mixing ratio

Problem
A solution contains $0.200 mol$ ethanol and $0.800 mol$ water. Find both mole fractions.

1.Total amount of component molecules is $0.200 + 0.800 = 1.000 mol$ .
2. $x_{ethanol} = 0.200/1.000 = 0.200$ and $x_{water} = 0.800$ .
3.The fractions add to 1. This does not mean the ethanol mass percentage is 20.0%, because ethanol and water have different molar masses.

Very dilute contamination is sometimes reported in parts per million by mass, abbreviated ppm by mass. It uses the same mass-ratio idea as mass percentage, but multiplies by one million rather than 100. Thus 1 ppm by mass is 1 mg solute per kilogram of solution; this extension helps interpret the chapter’s contamination exercise.

Parts per million by massLaTeX

ppm = \frac{m _{solute}}{m _{solution}} \times 1 0^{6}

The masses must use the same unit. A value in ppm by mass divided by

1 0^{4}

gives mass percent.

Molarity uses the final solution volume

For reactions, we often need moles of solute in a measured volume. Molarity gives that directly. Because the denominator is final solution volume, preparing a solution means dissolving the solute and then bringing the mixture to the specified total volume.

MolarityLaTeX

c = \frac{n _{solute}}{V _{solution}} n_{solute} = c V_{solution}

c

denotes molarity in

mol L^{- 1}

, often written M. Use litres for

V

in this form. Here

c

avoids confusing molarity with molar mass, also commonly symbolised M.

Example — Preparing sodium hydroxide solution

Problem
What is the molarity when $4.00 g$ NaOH is dissolved to make $250.0 mL$ solution? Use molar mass $40.00 g mo l^{- 1}$ .

1.Solute amount is $4.00/40.00 = 0.100 mol$ .
2.Final solution volume is $250.0 mL = 0.2500 L$ .
3. $c = 0.100/0.2500 = 0.400 mol L^{- 1}$ . The solution should be made up to 250.0 mL, rather than formed by adding the solid to 250.0 mL water.

Amount and concentration are different

$0.50 mol$ sodium carbonate is an amount. $0.50 mol L^{- 1}$ sodium carbonate describes a solution concentration. Its amount still depends on how much solution is taken.

Dilution preserves solute while changing concentration

A stock solution has a known, relatively high concentration. Adding solvent increases solution volume without adding solute, provided no solute is lost and no reaction changes it. The same number of solute moles is then distributed through a larger volume.

Dilution relationshipLaTeX

c_{1} V_{1} = c_{2} V_{2}

The subscripts identify the initial stock portion and the final diluted solution. Use matching volume units on both sides. This equation follows from equality of solute amounts, not from conservation of volume.

A fixed solute amount in a larger volume— Dots represent equal solute portions, not literal visible molecules. Add enough water to reach the final volume.

Example — Diluting a stock solution

Problem
Prepare $1.000 L$ of $0.200 mol L^{- 1}$ NaOH from a $1.00 mol L^{- 1}$ stock. What stock volume is needed?

1.The target contains $(0.200) (1.000) = 0.200 mol$ NaOH.
2.The stock portion must contain that amount: $V_{1} = 0.200/1.00 = 0.200 L = 200 mL$ .
3.Take 200 mL stock and add sufficient water to reach 1.000 L total solution. Do not assume that adding exactly 800 mL water necessarily sets the final volume accurately.

Molality uses solvent mass

Molality also counts solute moles, but divides by solvent mass in kilograms. This denominator is especially useful when temperature changes, because heating changes volume more directly than mass. Do not confuse the similar names molarity and molality.

MolalityLaTeX

b = \frac{n _{solute}}{m _{solvent} in kg}

b

is molality in

mol k g^{- 1}

, often denoted by a lower-case m. This lesson uses

b

to distinguish it from a mass symbol.

Example — Converting molarity to molality

Problem
A $3.00 mol L^{- 1}$ NaCl solution has density $1.25 g m L^{- 1}$ . Find its molality using NaCl molar mass $58.5 g mo l^{- 1}$ .

1.Choose a 1.00 L solution basis. It contains $3.00 mol$ NaCl, with mass $3.00 (58.5) = 175.5 g$ .
2.The solution mass is $1000 mL \times 1.25 g m L^{- 1} = 1250 g$ .
3.Solvent mass is solution mass minus solute mass: $1250 - 175.5 = 1074.5 g = 1.0745 kg$ .
4.Molality is $3.00/1.0745 = 2.79 mol k g^{- 1}$ to three significant figures. Retain the displayed working values until the final calculation.

Measure	Numerator	Denominator	Effect of temperature alone in a sealed nonreacting sample
Mass percentage	Solute mass	Solution mass	Unchanged if composition remains unchanged
Mole fraction	Component moles	Total component moles	Unchanged if composition remains unchanged
Molarity	Solute moles	Solution volume	Can change because volume changes
Molality	Solute moles	Solvent mass in kg	Unchanged if composition remains unchanged

Connecting concentration to a reaction

Molarity does not replace stoichiometry; it supplies the reactant amount. Multiply concentration by a volume in litres, use a balanced equation to relate amounts, and then convert the result if a mass is requested. This joins the solution description to the reaction model.

Example — An acid volume and a reacting solid

Problem
What mass of calcium carbonate reacts completely with $25.0 mL$ of $0.750 mol L^{- 1}$ HCl? Use $CaC O_{3} + 2HCl \to CaC l_{2} + C O_{2} + H_{2} O$ and $M (CaC O_{3}) = 100.0 g mo l^{- 1}$ .

1.The acid amount is $0.750 (0.0250) = 0.01875 mol$ .
2.The equation requires two moles HCl per mole carbonate, so carbonate amount is $0.01875/2 = 0.009375 mol$ .
3.Carbonate mass is $0.009375 (100.0) = 0.9375 g$ , reported as $0.938 g$ . A volume-to-mole step must precede the reaction ratio.

Quick check

The denominator in solute mass percentage is:

Quick check

Which concentration measure is dimensionless?

Quick check

How many moles are in $200 mL$ of $0.500 mol L^{- 1}$ solution?

Quick check

During dilution with pure solvent, which quantity stays fixed under the stated assumptions?

Quick check

Why can molarity change when a sealed solution is warmed?

Quick check

Which information is useful for converting molarity to molality?

Practice Problems

Find the mass percent when $5.00 g$ solute is mixed with $95.0 g$ water.
Find both mole fractions for $0.300 mol$ A mixed with $0.700 mol$ B.
Find the mass of sodium acetate needed for $500 mL$ of a $0.375 mol L^{- 1}$ solution using molar mass $82.0245 g mo l^{- 1}$ .
Find the molarity of nitric acid solution with density $1.41 g m L^{- 1}$ and acid content $69.0%$ by mass; use molar mass $63.0 g mo l^{- 1}$ .
What volume of $2.00 mol L^{- 1}$ stock is needed to make $250.0 mL$ of $0.200 mol L^{- 1}$ solution?
Convert $15 ppm$ by mass to mass percent. For 1.000 kg of solution, identify solute and solvent masses before finding molality using molar mass $119.37 g mo l^{- 1}$ .
Explain why the mass of the solvent, rather than the whole solution, appears in molality.

Key Takeaways

Each concentration measure has a specific denominator; solution and solvent are not interchangeable. Mass percentage uses solution mass, while mole fraction uses total component moles. Molarity is solute moles per litre of final solution; molality is solute moles per kilogram of solvent. Dilution conserves solute amount when no loss or reaction occurs. Use density to connect solution volume to mass, and a balanced equation to connect reacting amounts.

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Chemical Equations, Stoichiometry, and Limiting Reagents

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Chapter Summary and Practice

How much solute is present?

Mass percentage and mole fraction

Molarity uses the final solution volume

Dilution preserves solute while changing concentration

Molality uses solvent mass

Connecting concentration to a reaction

Quiz

Practice Problems

Key Takeaways