Question:
A gas mixture contains \(64\%\) methane and \(36\%\) ethane by mass. The density of the mixture at \(27^\circ\mathrm{C}\) and \(750\ \mathrm{mmHg}\) pressure (in \(\mathrm{g\,L^{-1}}\)) is
\[
(\text{At. wt. of C}=12,\; \text{H}=1,\; R=0.082\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}})
\]
A gas mixture contains \(64\%\) methane and \(36\%\) ethane by mass. The density of the mixture at \(27^\circ\mathrm{C}\) and \(750\ \mathrm{mmHg}\) pressure (in \(\mathrm{g\,L^{-1}}\)) is
\[
(\text{At. wt. of C}=12,\; \text{H}=1,\; R=0.082\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}})
\]
Show Hint
For gaseous mixtures:
\[
\boxed{d=\frac{PM}{RT}}
\]
First calculate the average molar mass using the given composition, then substitute into the gas density formula.
Updated On: Jul 9, 2026
- 0.57
- 0.87
- 0.67
- 0.77 \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Density of a gas mixture is given by
\[
d=\frac{PM}{RT}
\]
where \(M\) is the average molar mass of the gas mixture.
Step 1: Calculate the average molar mass. Assume \(100\,\mathrm{g}\) of the mixture. \[ \begin{aligned} \text{Moles of CH}_4 &=\frac{64}{16}=4 \text{Moles of C}_2\text{H}_6 &=\frac{36}{30}=1.2 \end{aligned} \] Total moles \[ 4+1.2=5.2 \] Average molar mass \[ M=\frac{100}{5.2}=19.23\ \mathrm{g\,mol^{-1}} \]
Step 2: Calculate the density. Pressure \[ P=\frac{750}{760}=0.987\ \mathrm{atm} \] Temperature \[ T=27+273=300\ \mathrm{K} \] \[ d=\frac{0.987\times19.23}{0.082\times300} =0.77\ \mathrm{g\,L^{-1}} \]
Step 3: Final conclusion. \[ \boxed{0.77\ \mathrm{g\,L^{-1}}} \] Hence, the correct option is \(\boxed{(D)}\).
Step 1: Calculate the average molar mass. Assume \(100\,\mathrm{g}\) of the mixture. \[ \begin{aligned} \text{Moles of CH}_4 &=\frac{64}{16}=4 \text{Moles of C}_2\text{H}_6 &=\frac{36}{30}=1.2 \end{aligned} \] Total moles \[ 4+1.2=5.2 \] Average molar mass \[ M=\frac{100}{5.2}=19.23\ \mathrm{g\,mol^{-1}} \]
Step 2: Calculate the density. Pressure \[ P=\frac{750}{760}=0.987\ \mathrm{atm} \] Temperature \[ T=27+273=300\ \mathrm{K} \] \[ d=\frac{0.987\times19.23}{0.082\times300} =0.77\ \mathrm{g\,L^{-1}} \]
Step 3: Final conclusion. \[ \boxed{0.77\ \mathrm{g\,L^{-1}}} \] Hence, the correct option is \(\boxed{(D)}\).
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