Question

A concentrated solution of sulphuric acid is 98% $H_{2}S{O}_4$ by mass and has a density of $1.84g/mL$. What volume of this concentrated acid solution is required to make $6L$ of a $6.13M{H}_{2}S{O}_4$ solution?
(Molar mass of $H_{2}S{O}_4=98g/mol$)

• A. $6.5L$
• B. $3.2L$
• C. $2L$
• D. $4.4L$

Answer 1

The correct option is C $2L$

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,
$(18.4\times V)=(6.13\times 6)$
$V=2L$
Volume of 18.4 M $H_{2}S{O}_4$ required is $2L$.