Question

A gaseous mixture contains SO3(g) and CH. (g) in 25: 1 ratio by mass. What is ratio of total number of atoms present in SO3(g) to total number of atoms present in CH4 (g) in the mixture.​

Answer 1

Answer:

Ratio of total number of atoms present in SO3(g) to total number of atoms present in CH4 (g) in the mixture is 2:1

Explanation:

We will solve by assuming some mass to the gas mixture.

Let the mass of oxygen $\mathrm{O}_2$ be 1g} in the mixture, then according to question, mass of sulphur dioxide $\mathrm{SO}_2$ is 4g

Now, for $\mathrm{O}_2$

1 mole of $O_2$ atoms $=6.022 \times 10^{23}$ atoms

Now, as we know, molecular mass of $O_2=32 \mu$

(mass number of oxygen is $16 \mu$, thus molecular mass $=16+16=32$ )

Thus, molar mass of $O_2=32 \mathrm{~g}$

$\begin{array}{llll} & \mathrm{SO}_3 & : & \mathrm{CH}_4 \\\text { Mass ratio } & 12.5 & : & 1 \\\text { Mass ratio } & \frac{12.5}{80} & : & \frac{1}{16} \rightarrow 2.5: 1 \\\text { Atom ratio } & 2.5 \times 4: 5 \times 1 & \Rightarrow 2: 1\end{array}$

Therefore, Ratio of total number of atoms present in SO3(g) to total number of atoms present in CH4 (g) in the mixture is 2:1

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