Ethanol (C₂H₆O) and methanol (CH₄O) form nearly ideal mixtures. A mixture consisting of 1.02 moles of ethanol and 0.98 moles of methanol has a total vapour pressure of 8.82 kPa at 293 K. When another 0.5 moles of methanol is added, the vapour pressure rose to 9.32 kPa. The vapour pressures of pure methanol and ethanol respectively (in bar) are

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0.128 and.0.226

0.226 and 0.118

0.064 and 0.113

0.113 and 0.064

answer is B.

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Detailed Solution

(A)
$\begin{array}{l} 8 .82 = p_{{CH}_{3} OH}^{\circ} \times \frac{0 .98}{2} + p_{C_{2} H_{5} OH}^{\circ} \times \frac{1 .02}{2} \\ = 0 .49 \times p_{{CH}_{3} OH}^{\circ} + 0 .51 p_{C_{2} H_{5} OH}^{\circ} \end{array}$
(B)
$\begin{array}{l} 9 .32 = p_{{CH}_{3} OH}^{\circ} \times \frac{1 .48}{2 .5} + p_{C_{2} H_{5} OH}^{\circ} \times \frac{1 .02}{2 .5} \\ = 0 .592 \times p_{{CH}_{3} OH}^{\circ} + 0 .408 p_{C_{2} H_{5} OH}^{\circ} \end{array}$
(C) On solving the two equations
$\begin{array}{l} p_{C_{2} H_{5} OH}^{\circ} = 6 .41 kPa = 6.41 kPa \times 0.01 \frac{bar}{kPa} = 0.064 bar \\ p_{{CH}_{3} OH}^{\circ} = 11 .33 kPa \end{array}$
or $p_{{CH}_{3} OH}^{\circ} = 11.33 kPa \times 0.01 \frac{bar}{kPa} = 0.113 bar$