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Car batteries use a solution of sulfuric acid in water…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Car batteries use a solution of sulfuric acid in water. A teacher has a bottle of battery acid labelled $39.2\%$ $\text{H}_2\text{SO}_4$ by mass, with a density of $1.30\ \text{g mL}^{-1}$ . She wants to use it for three class experiments at room temperature, and first needs its concentration in the units that NCERT uses. A student says that the molarity and the molality must be close, because the solvent is water. Your answers to (a) and (b) will test that claim. For safety, the acid is always poured slowly into the water, never the water into the acid; the order does not change any mass or amount below. Atomic masses in $\text{g mol}^{-1}$ : H 1, O 16, S 32.

What is the molality of the battery acid?
What is its molarity?
What is the mole fraction of $\text{H}_2\text{SO}_4$ in it?
What volume of the battery acid must be diluted with water to make exactly $500\ \text{mL}$ of $0.260\ \text{M}$ sulfuric acid?
What mass of water must be added to $100\ \text{g}$ of the battery acid to make an acid that is $9.80\%$ $\text{H}_2\text{SO}_4$ by mass?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $39.2\%$ $\text{H}_2\text{SO}_4$ by mass, density $1.30\ \text{g mL}^{-1}$ . Molar masses: $\text{H}_2\text{SO}_4$ $98\ \text{g mol}^{-1}$ , water $18\ \text{g mol}^{-1}$ .

Plan. Take $100\ \text{g}$ of the acid: $39.2\ \text{g}$ of $\text{H}_2\text{SO}_4$ , that is $0.400\ \text{mol}$ , and $60.8\ \text{g}$ of water, that is $3.378\ \text{mol}$ .

Step 1. Molality. $m = \dfrac{0.400\ \text{mol}}{0.0608\ \text{kg}} = 6.58\ \text{mol kg}^{-1}$ .

Step 2. Molarity. The volume of $100\ \text{g}$ is $\dfrac{100\ \text{g}}{1.30\ \text{g mL}^{-1}} = 76.92\ \text{mL}$ . Then $M = \dfrac{0.400\ \text{mol}}{0.07692\ \text{L}} = 5.20\ \text{mol L}^{-1}$ . The student's claim fails: the two differ by more than one unit, because this solution is far from dilute.

Step 3. Mole fraction. $x = \dfrac{0.400}{0.400 + 3.378} = 0.106$ .

Step 4. Dilution by volume. The moles of acid needed are $0.260\ \text{mol L}^{-1} \times 0.500\ \text{L} = 0.130\ \text{mol}$ . We want the volume of stock, so $V_1 = \dfrac{0.130\ \text{mol}}{5.20\ \text{mol L}^{-1}} = 25.0\ \text{mL}$ .

Step 5. Dilution by mass. The $39.2\ \text{g}$ of acid must be $9.80\%$ of the new mass: $\dfrac{39.2\ \text{g}}{0.0980} = 400\ \text{g}$ in all. Water to add: $400\ \text{g} - 100\ \text{g} = 300\ \text{g}$ .

Answer: (a) $6.58\ \text{mol kg}^{-1}$ . (b) $5.20\ \text{mol L}^{-1}$ . (c) $0.106$ . (d) $25.0\ \text{mL}$ . (e) $300\ \text{g}$ .

Check. Take one litre instead: mass $1300\ \text{g}$ , of which $39.2\%$ is acid, $5.20\ \text{mol}$ , the molarity found. The water in it is $790.4\ \text{g}$ , and $\dfrac{5.20\ \text{mol}}{0.7904\ \text{kg}} = 6.58\ \text{mol kg}^{-1}$ , the molality found.

Watch out. Molality divides by the mass of the water ( $60.8\ \text{g}$ ), not of the whole solution ( $100\ \text{g}$ ). In (e), adding $300\ \text{g}$ of water makes the mass four times larger, so the per cent falls to a quarter.

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Car batteries use a solution of sulfuric acid in water…

The worked solution

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