Solution Concentration

20min Part 1 / Ch2 / Lesson 2

Prerequisites: 1-2-1

Objectives

Convert between mass percent concentration and molarity
Perform dilution calculations
Use concentration units correctly

Two Ways to Express Concentration

Mass percent vs. molarity

Mass percent concentration $= \dfrac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 100$ (%)

Molarity $c = \dfrac{n\ \text{(mol)}}{V\ \text{(L)}}$ (mol/L)

Converting Mass Percent → Molarity

Given density $d\ \mathrm{(g/mL)}$ , mass percent $w$ (%), and molar mass $M\ \mathrm{(g/mol)}$ :

c = \frac{10dw}{M}\ \ \mathrm{(mol/L)}

Derivation: $1\ \mathrm{L} = 1000\ \mathrm{mL}$ of solution has mass $1000d\ \mathrm{g}$ . The solute in it is $1000d \times \frac{w}{100} = 10dw\ \mathrm{g}$ . In moles: $\frac{10dw}{M}\ \mathrm{mol}$ .

Dilution

Dilution does not change the amount of solute: $c_1 V_1 = c_2 V_2$

Worked Example

Example: Mass Percent → Molarity

Find the molarity of hydrochloric acid with density $1.20\ \mathrm{g/mL}$ and mass percent $36.5\%$ . ( $\mathrm{HCl} = 36.5$ )

$c = \frac{10 \times 1.20 \times 36.5}{36.5}$

$c = \frac{10 \times 1.20 \times 36.5}{36.5} = 10 \times 1.20 = 12.0\ \mathrm{mol/L}$

Check Your Understanding

Q1 20 g of salt is dissolved in 180 g of water. What is the mass percent?

Q2 How many moles of NaOH are in 200 mL of 0.50 mol/L NaOH solution?

Q3 100 mL of 2.0 mol/L solution is diluted to 500 mL. What is the new molarity?

Exercises

Q1. You need to prepare $500\ \mathrm{mL}$ of $0.20\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$ . How many mL of concentrated sulfuric acid ( $98\%$ , density $1.84\ \mathrm{g/mL}$ , $M = 98.0$ ) are needed?

Solution

Moles of $\mathrm{H_2SO_4}$ needed: $n = 0.20 \times 0.500 = 0.10\ \mathrm{mol}$

Molarity of concentrated sulfuric acid: $c = \frac{10 \times 1.84 \times 98}{98.0} = 18.4\ \mathrm{mol/L}$

Volume of concentrated sulfuric acid needed: $V = \frac{n}{c} = \frac{0.10}{18.4} = 5.43 \times 10^{-3}\ \mathrm{L} = 5.4\ \mathrm{mL}$