Question

A colourless salt (X) has $50\%$ $N{a}_{2}S{O}_3$ and $50\%H_{2}O$. Then (a)The formula of (X) is
(b)The amount of $S{O}_2$ at NTP obtained when 2.52 g of (X) reacts with excess of dil. $H_{2}S{O}_4$ is_________.

• A. $N{a}_{2}S{O}_3\cdot 7H_{2}O$ , 0.224 L at NTP from 2.52 g of (X)
• B. $N{a}_{2}S{O}_3\cdot 6H_{2}O$ , 0.224 L at NTP from 2.52 g of (X)
• C. $N{a}_{2}S{O}_3\cdot 7H_{2}O$ , 2.24 L at NTP from 2.52 g of (X)
• D. $N{a}_{2}S{O}_3\cdot 6H_{2}O$ ,2.24 L at NTP from 2.52 g of (X)

Answer 1

Answer: (A)

$N{a}_{2}S{O}_3\cdot 7H_{2}O$ , 0.224 L at NTP from 2.52 g of (X)

(X) is $N{a}_{2}S{O}_3\cdot 7H_{2}O$
$N{a}_{2}S{O}_3\cdot 7H_{2}O+H_{2}S{O}_4\to N{a}_{2}S{O}_4+S{O}_2+8H_{2}O$
$\begin{array}{c}N{a}_{2}S{O}_3\cdot 7H_{2}O\\ 252g\end{array}+H_{2}S{O}_4\to N{a}_{2}S{O}_4+\begin{array}{c}S{O}_2\uparrow \\ 22.4L\\ atNTP\end{array}+8H_{2}O$
So here 252 grams of $N{a}_{2}S{O}_3\cdot 7H_{2}O$ gave 22.4 liters of $O_2$ at NTP
2.52 grams of $N{a}_{2}S{O}_3\cdot 7H_{2}O$ will give how many liters of $O_2$ at NTp
$S{O}_2:0.224L$ at NTP from $2.52gN{a}_{2}S{O}_3\cdot 7H_{2}O$.
Hence option A is correct.