# Question 33e2d

Jul 9, 2017

a) Molarity = 14 mol/L; b) volume of acid = 18 mL.

#### Explanation:

a) Molarity of acid

Assume that you have 1 L of the solution.

$\text{Mass of acid" = 1000 color(red)(cancel(color(black)("mL solution"))) × "1.73 g solution"/(1 color(red)(cancel(color(black)("mL solution")))) = "1730 g solution}$

${\text{Mass of H"_2"SO"_4 = 1730 color(red)(cancel(color(black)("g solution"))) × ("80 g H"_2"SO"_4)/(100 color(red)(cancel(color(black)("g solution")))) = "1380 g H"_2"SO}}_{4}$

${\text{Moles of H"_2"SO"_4 = 1380 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 "mol H"_2"SO"_4)/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "14.1 mol H"_2"SO}}_{4}$

$\text{Molarity" = "moles"/"litres" = "14.1 mol"/"1 L" = "14 mol/L}$

Note: The answer can have only 2 significant figures, because that is all you gave for the percentage composition of the acid.

b) Volume of acid

To solve this problem, we can use the dilution formula

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange the formula to get

V_1 = V_2 × c_2/c_1

In this problem,

c_1 = "14.1 mol/L"; V_1 = ?#
${c}_{2} = \text{0.50 mol/L"; V_2 = "0.500 L}$

${V}_{1} = \text{0.500 L" × (0.50 color(red)(cancel(color(black)("mol/L"))))/(14.1 color(red)(cancel(color(black)("mol/L")))) = "0.018 L" = "18 mL}$

You would need to use 18 mL of the 80 % acid.