Question

1.80 g of a metal oxide required 833 mL of hydrogen at NTP to be reduced to its metal. Find the equivalent weight of the oxide and the metal.

• A. 25.67 and 28.92
• B. 24.5 and 16.2
• C. 28.21 and 12.25
• D. None of these

Answer 1

The correct option is B 24.5 and 16.2

$vol=833ml$ at NTP

metal at NTP $=\frac{833\times {10}^{-3}}{22.7}$

$\approx 36.69\times {10}^{-3}mol$

$36.69\times {10}^{-3}$ metal $H_2$ reacts with $1.80g$ and

$=\frac{1.80}{36.69\times {10}^{-3}}$

$\approx 49.05$

$\therefore$ Eq wt of oxide=$\frac{mass}{n-factor}$

$=\frac{49.05}{2}$

$\approx 24.5$

Equation of metal. Equation of article -Equation of oxygen

$=24.5-8$

$=16.5$