Question

$ 0.32\, g $ of metal gave on treatment with an add $ 112\,mL $ of hydrogen at $ NTP $ . Calculate the equivalent weight of the metal.

• A. $ 58 $
• B. $ 32 $
• C. $ 11.2 $
• D. $ 24 $

Answer 1

Answer: (B)

Eq. weight of metal
$=\frac{\text{ weight of metal}}{\text{ weight of hydrogen displaced}}\times1.008 $
weight of hydrogen liberated $=\frac{112}{22400}\times 2.016$
$=1.008\times10^{-2}$
$\therefore $ Eq. weight of metal $=\frac{0.32}{1.008\times 10^{-2}} \times 1.008$
$=32\,g\, equiv^{-1}$