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\(2 .9 \mathrm{~g}\) of a gas at \(95^{\circ} \mathrm{C}\) occupied the same volume as
\(0.184\) g of dihydrogen at \(17^{\circ} \mathrm{C}\), at the same pressure.
What is the molar mass of the gas ?
\(0.184\) g of dihydrogen at \(17^{\circ} \mathrm{C}\), at the same pressure.
What is the molar mass of the gas ?
Solution:
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Verified Answer
Let the molar mass of the gas be \(\mathrm{M}_{\mathrm{x}}\)
We know \(\mathrm{P}_1=\mathrm{P}_2\) and \(\mathrm{V}_1=\mathrm{V}_2\)
\(\begin{array}{ll}\text { or } & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ \text { or } & \mathrm{n}_1 \mathrm{RT}_1=\mathrm{n}_2 \mathrm{RT}_2 \quad \text { or } \quad \mathrm{n}_1 \mathrm{~T}_1=\mathrm{n}_2 \mathrm{~T}_2\end{array}\)
\(\begin{aligned}
&\text { or } \frac{\mathrm{W}_1}{\mathrm{M}_1} \mathrm{~T}_1=\frac{\mathrm{W}_2}{\mathrm{M}_2} \mathrm{~T}_2 \\
&\therefore \frac{2.9}{\mathrm{M}_{\mathrm{x}}} \times(95+273)=\frac{0.184}{2} \times(17+273)
\end{aligned}\)
or \(\mathrm{M}_{\mathrm{x}}=\frac{2.9 \times 368 \times 2}{0.184 \times 290}=40 \mathrm{~g} \mathrm{~mol} \mathrm{~mol}^{-1}\)
We know \(\mathrm{P}_1=\mathrm{P}_2\) and \(\mathrm{V}_1=\mathrm{V}_2\)
\(\begin{array}{ll}\text { or } & \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 \\ \text { or } & \mathrm{n}_1 \mathrm{RT}_1=\mathrm{n}_2 \mathrm{RT}_2 \quad \text { or } \quad \mathrm{n}_1 \mathrm{~T}_1=\mathrm{n}_2 \mathrm{~T}_2\end{array}\)
\(\begin{aligned}
&\text { or } \frac{\mathrm{W}_1}{\mathrm{M}_1} \mathrm{~T}_1=\frac{\mathrm{W}_2}{\mathrm{M}_2} \mathrm{~T}_2 \\
&\therefore \frac{2.9}{\mathrm{M}_{\mathrm{x}}} \times(95+273)=\frac{0.184}{2} \times(17+273)
\end{aligned}\)
or \(\mathrm{M}_{\mathrm{x}}=\frac{2.9 \times 368 \times 2}{0.184 \times 290}=40 \mathrm{~g} \mathrm{~mol} \mathrm{~mol}^{-1}\)
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