Question

A gaseous mixture contains ${CH}_4$ and ${CH}_2{H}_6$ in equimolecular proportion. The weight of $2.24$ liters of this mixture at NTP is:

• A. $4.6g$
• B. $1.6g$
• C. $2.3g$
• D. $23g$

Answer 1

The correct option is C $2.3g$

Equal no. of molecules contained by equal volumes.

$\therefore$ Volume of $C{H}_4=$ Volume of $C_2{H}_6=V$ (Assumed)

Given that:- total volume of mixture $=2.24L$

$\therefore$ Volume of $C{H}_4$ + Volume of $C_2{H}_6=2.24$

$\Rightarrow V+V=2.24$

$\Rightarrow 2V=2.24$

$\Rightarrow V=1.12L$

$\therefore$ Volume of $C{H}_4$ = Volume of $C_2{H}_6=V=1.12L$

Mol. wt. of $C{H}_4=16g$

Mol. wt. of $C_2{H}_6=30g$

$\therefore$ weight of 1.12 L of $C{H}_4=\frac{16}{22.4}\times 1.12=0.8g$

weight of 1.12 L of $C_2{H}_6=\frac{30}{22.4}\times 1.12=1.5$

$\therefore$ Total weight of mixture = wt. of $C{H}_4+$ wt. of $C_2{H}_6=0.8+1.5=2.3g$

Hence, the weight of 2.24 liters of this mixture at NTP is 2.3 g.