Question

$0.05$ g of a piece of metal in dilute acid gave $24.62$ mL of $H_2$ at ${27}^{o}C$ and $760$ mm pressure. The equivalent weight of the metal is :

• A. 25
• B. 12.5
• C. 50
• D. 37.5

Answer 1

Answer: (A)

25

Volume of $H_2$ at STP $=24.62\times \frac{273}{300}=22.40$ mL
$22400$ mL of $H_2$ at STP $=1$mole $=2$ Eq. of $H_2$
$22.4$ mL of $H_2=\frac{2}{22400}\times 22.4$ $=0.002$ Eq of $H_2$ $=0.002$ Eq of metal
$\therefore \frac{Weight}{EW}=0.002$
$\frac{0.05}{0.002}=EW=25$