Question

The weight of mixture containing HCl and $H_{2}S{O}_4$ is $0.1$g on treatment with an excess of an $AgN{O}_3$ solution, reacted with this acid mixture gives $0.1435$g of AgCl, weight percentage of $H_{2}S{O}_4$ in mixture is:

• A. $36.5$
• B. $63.5$
• C. $50$
• D. $80$

Answer 1

Answer: (B)

$63.5$

Let mass of $HCl$ be $x$

$\therefore$ Mass of $H_{2}S{O}_4$ is $0.1-x$

Moles of $HCl=\frac{x}{36.5}$

$HCl+AgN{O}_3⟶AgCl\downarrow +HN{O}_3$

Moles of $HCl=$ moles of $AgCl$

Moles of $AgCl=\frac{0.1435}{143.3}={10}^{-3}$ moles

$\therefore \frac{x}{36.5}={10}^{-3}$

$x=36.5\times {10}^{-3}$

Weight of $HCl=0.0365$ $g$

Weight of $H_{2}S{O}_4=0.0635$

$\%$ weight of $H_{2}S{O}_4=\frac{0.0635}{0.1}\times 100=63.5\%$