Question

If the weight of 5.6 lit of a gas at NTP is 11g, the gas may be: (at. wt. of $P=31,N=14.O=16$ and $Cl=35.5)$.

• A. phosphine
• B. phosgene
• C. nitric oxide
• D. nitrous oxide

Answer 1

The correct option is D nitrous oxide

Mass of the gas $m=11gm$

Volume occupied by the gas

$=V=5.6l$

$NTP\Rightarrow$ temperature of the gas $=T=273K$ and pressure exerted by the gas $=P=1atm$

$Gas=W=?$

[$\because$ any gas can be identified by its molecular weight]

$PV=\frac{m}{W}RT$

$\Rightarrow W=\frac{mRT}{PV}.....\left(1\right)$

Substituting the values from the data in equation (1), we get

$W=\frac{11\times 0.0821\times 273}{1\times 5.6}$

$\Rightarrow W=\frac{11\times 22.4}{5.6}$

$\Rightarrow W=11\times 4=44$

Therefore, the gas can be either carbon dioxide, $C{O}_2$ [or] nitrous oxide, $N_{2}O$ [or] propane, $C_3{H}_3$.