Question

When $1\times {10}^{-3}$ mol of the chloride of an element $Y$ was completely hydrolysed, it was found that the resulting solution required $20$ mL of $0.1M$ aqueous silver nitrate for complete precipitation of the chloride ion. Element $Y$ could be:

• A. aluminium
• B. phosphorous
• C. silicon
• D. sulphur

Answer 1

The correct option is D sulphur

$YC{l}_n+n{H}_{2}O\to nHCl+Y(OH{)}_n$

($n=$ valency of $Y$)

$nHCl+nAgN{O}_3\to nAgCl$

$\frac{1}{n}=\frac{1\times {10}^{-3}}{20\times 0.1\times {10}^{-3}}$ $\left(\frac{Molarity\times Volume}{1000}=\text{number of mole}\right)$

$n=2$

Therefore, the element must be sulphur (others have valency, $n>2$).