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Standard XII

Chemistry

Liquids in Liquids

If 2 moles of...

Question

If 2 moles of A and 3 moles of B are mixed to form an ideal solution vapour pressure of A and B are 120 and 180 mm of Hg respectively. then the composition of A and B in the Vapour phase when the first traces of vapour are formed in the above case is:

X^{l} A = 0.407

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X^{l} A = 0.8

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X^{l} A = 0.109

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X^{l} A = 0.307

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Solution

The correct option is D $X^{l} A = 0.307$
$X_{A} = \frac{2}{2 + 3} = \frac{2}{5}$
$X_{B} = \frac{3}{2 + 3} = \frac{3}{5}$
$X_{A}^{1} = \frac{P_{A}^{0} \times A}{P_{t o t a l}}$
$= 120 \times \frac{2}{5}$
$= 120 \times \frac{2}{5} + 180 \times \frac{3}{5}$
$= 48$
$= 48 + 108$
$X_{A}^{1} = 0.307$

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If 2 moles of A and 3 moles of B are mixed to form an ideal solution vapour pressure of A and B are 120 and 180 mm of Hg respectively. then the composition of A and B in the Vapour phase when the first traces of vapour are formed in the above case is:

The correct option is D XlA=0.307XA=22+3=25XB=32+3=35X1A=P0A×APtotal =120×25 =120×25+180×35 =48 =48+108X1A=0.307

The correct option is D $X^{l} A = 0.307$
$X_{A} = \frac{2}{2 + 3} = \frac{2}{5}$
$X_{B} = \frac{3}{2 + 3} = \frac{3}{5}$
$X_{A}^{1} = \frac{P_{A}^{0} \times A}{P_{t o t a l}}$
$= 120 \times \frac{2}{5}$
$= 120 \times \frac{2}{5} + 180 \times \frac{3}{5}$
$= 48$
$= 48 + 108$
$X_{A}^{1} = 0.307$