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In a flask, the weight ratio of $\mathrm{CH}_{4}(g)$ and $\mathrm{SO}_{2}(g)$ at $298 \mathrm{K}$ and 1 bar is $1: 2 .$ The ratio of the number of molecules of $\mathrm{SO}_{2}(\mathrm{g})$ and $\mathrm{CH}_{4}(g)$ is
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$1 : 2$
Given, weight ratio : $W_{\mathrm{C H_{4}},}: W_{\mathrm{SO}_{2}}=1: 2$ $\because \quad n=\frac{w}{m} \quad\left[\begin{array}{l}n=\text { no. of mole } \\ w=\text { mass } \\ M=\text { molar mas: }\end{array}\right.$
$\therefore \quad \frac{n_{1}}{n_{2}}=\frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}\left\{\begin{array}{l}n_{1}=n_{\mathrm{so}_{1}} \\ n_{2}=n_{\mathrm{Oi}_{2}}\end{array}\right.$
$=\frac{2}{64} \times \frac{16}{1}$
$\frac{n_{1}}{n_{2}}=\frac{1}{2},$ i.e. 1: 2
Also $n \propto N$
$\therefore$ Ratio of number of molecules is 1: 2.
$\therefore \quad \frac{n_{1}}{n_{2}}=\frac{W_{1}}{M_{1}} \times \frac{M_{2}}{W_{2}}\left\{\begin{array}{l}n_{1}=n_{\mathrm{so}_{1}} \\ n_{2}=n_{\mathrm{Oi}_{2}}\end{array}\right.$
$=\frac{2}{64} \times \frac{16}{1}$
$\frac{n_{1}}{n_{2}}=\frac{1}{2},$ i.e. 1: 2
Also $n \propto N$
$\therefore$ Ratio of number of molecules is 1: 2.
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