Question

$2.56 \times 10^{-3}$ equivalent of $KOH$ is required to neutralise $0.12544\, g\, H _{2} XO _{4}$. The atomic mass of $X$ (in $g / mol$ ) is: [Given: $H _{2} X O _{4}$ is a dibasic acid]

• A. $16$
• B. $8$
• C. $7$
• D. $32$

Answer 1

Answer: (D)

Let the atomic weight of $X = y$
No. of equivalents of $KOH =$ No. of equivalents of $H _{2} XO _{4}$
i.e. $\frac{0.12544}{\frac{66+y}{2}}=2.56\times 10^{-3}$
$\therefore y =32$