Question

A concentrated solution of sulphuric acid is 98% $H_{2}S{O}_4$ by mass and has a density of 1.84 g/mL. What volume of this concentrated acid solution is required to make 5 L of a 9.2 M $H_{2}S{O}_4$ solution?
(Molar mass of $H_{2}S{O}_4$ = 98 g/mol)

• A. 6.5 L
• B. 1.7 L
• C. 2.5 L
• D. 3.2 L

Answer 1

The correct option is C 2.5 L

Let the mass of the $H_{2}S{O}_4$ solution be $100g$.
98% by mass means 98 g of $H_{2}S{O}_4$ is present in 100 g of solution.
$\text{Number of moles of}{H}_{2}S{O}_4=\frac{\text{given mass}}{\text{molar mass}}$
$=\frac{98}{98}=1$

We know, $\text{Density}=\frac{\text{mass}}{\text{volume}}$
So, $\text{volume of the solution}=\frac{100}{1.84}mL$

$\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution in mL}}\times 1000$

$\text{Molarity}=\frac{1}{\frac{100}{1.84}}\times 1000=18.4M$
Let the volume of 18.4 M acid be V L.
So, conserving moles before and after dilution:
$M_1{V}_1=M_2{V}_2$
where, $M_2$ and $V_2$ are the final concentration and volume,
$(18.4\times V)=(9.2\times 5)$
$V=2.5L$
Volume of 18.4 M $H_{2}S{O}_4$ required is 2.5 L.