Question

The atomic weight of zinc is $65.4$. The volume of dry hydrogen at NTP liberated by treating $6.54g$ of zinc with dilute sulphuric acid would be:

• A. $22.4$ litres
• B. $11.2$ litres
• C. $2.24$ litres
• D. $1.12$ litres

Answer 1

Answer: (C)

$2.24$ litres

$Zn+H_{2}S{O}_4\to ZnS{O}_4+H_2$

The volume of 1 mol of gas at N.T.P. is 22.4 litres. Moles of zinc used is equal to moles of hydrogen liberated.

Moles of zinc used is $\frac{6.54}{65.4}$=0.1 mol

Thus volume of hydrogen formed is $0.1\times 22.4$=$2.24$ litres