Question

What is the purity of concentrated $H_{2}S{O}_4$ solution $(d=1.8gm/mol)$ if $5\text{ml}$ of these solution is neutralized by $84.5\text{ml}$ of $2N\text{NaOH}$ solution.

• A. $93\text{\%}$
• B. $94.6\text{\%}$
• C. $92.12\text{\%}$
• D. $91.5\text{\%}$

Answer 1

The correct option is C $92.12\text{\%}$

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
Moles of $NaOH$ required $=\frac{2\times 84.5}{1000}=0.169$
For $2molesNaOH\to 1mole{H}_{2}S{O}_4$ is used
For $0.169moleNaOH\to 0.0845moles$ are used.
Mass of $H_{2}S{O}_4\to 0.0845\times 98=8.281gm$ (Theoretical)
Mass of $H_{2}S{O}_4$ used in original reaction $\Rightarrow 5\times 1.8=9gm$
$\therefore$ % purity $=\frac{8.281}{9}\times 100=92.12$ %