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. 0.57
• B. 0.87
• C. 0.67
• D. 0.77 \bigskip

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.

Answer 1

The Correct Option is D

Concept: Density \(d = PM/RT\). Find average molar mass from mass percentages.

Step 1: 64% CH₄, 36% C₂H₆. Per 100g: moles CH₄=64/16=4, C₂H₆=36/30=1.2. Total moles=5.2. M=100/5.2=19.23 g/mol. P=750/760=0.987 atm, T=300 K. \(d = 0.987\times19.23/(0.082\times300) = 0.77\) g/L.

Step 2: Write the final answer. \(\boxed{0.77\ \mathrm{g\,L^{-1}}}\)