Question

In a flask, the weight ratio of $CH _{4}( g )$ and $SO _{2}(g)$ at $298 \,K$ and 1 bar is $1: 2$. The ratio of the number of molecules of $SO _{2}(g)$ and $CH _{4}(g)$ is

• A. $1 : 4$
• B. $4 : 1$
• C. $1 : 2$
• D. $2 : 1$

Answer 1

Answer: (C)

$1 : 2$

Answer 2

Given, weight ratio : $W_{ CH _{4}}: W_{ sO _{2}}=1: 2$
$\because n=\frac{W}{m}$
$n=$ no. of moles
$w=$ mass
$M=$ molar mass
$\therefore \frac{n_{1}}{n_{2}}=\frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}\begin{cases}n_{1}=n_{ so _{2}} \\\ n_{2}=n_{ cH _{4}}\end{cases}$
$\frac{n_{1}}{n_{2}}=\frac{W_{ sO _{2}}}{M_{ sO _{2}}} \times \frac{M_{ CH _{4}}}{W_{ CH _{4}}}$
$=\frac{2}{64} \times \frac{16}{1}$
$\frac{n_{1}}{n_{2}}=\frac{1}{2}$, i.e. $1: 2$
Also $n \propto N$
$\therefore$ Ratio of number of molecules is $1: 2$.