In a flask, the weight ratio of CH_{4(g)} and | vedclass

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Question

(C) Given weight ratio of $CH_4$ and $SO_2$ is $1:2$. Let the weight of $CH_4 = w$ and weight of $SO_2 = 2w$. Molar mass of $CH_4 (M_1) = 16 \ g/mol$.

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(C) Given weight ratio of $CH_4$ and $SO_2$ is $1:2$. Let the weight of $CH_4 = w$ and weight of $SO_2 = 2w$. Molar mass of $CH_4 (M_1) = 16 \ g/mol$. Molar mass of $SO_2 (M_2) = 64 \ g/mol$. Number of moles of $CH_4 (n_1) = \frac{w}{16}$. Number of moles of $SO_2 (n_2) = \frac{2w}{64} = \frac{w}{32}$. The ratio of the number of molecules is equal to the ratio of the number of moles. Ratio $= \frac{n_{SO_2}}{n_{CH_4}} = \frac{w/32}{w/16} = \frac{16}{32} = \frac{1}{2}$. Thus, the ratio is $1:2$.

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

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