JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 15)
Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7 $$ \times $$ 103 Pa and 12 $$ \times $$ 103 Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temperature is :
xA = 0.76; xB = 0.24
xA = 0.28; xB = 0.72
xA = 0.4; xB = 0.6
xA = 0.37; xB = 0.63
Explanation
$${y_A} = {{{P_A}} \over {{P_{Total}}}} = {{P_A^0{x_A}} \over {P_A^0{X_A} \times P_B^0X{}_B}}$$
$$ = {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}$$
$$ = {{2.8} \over {10}} = 0.28$$
$${y_B} = 0.72$$
$$ = {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}$$
$$ = {{2.8} \over {10}} = 0.28$$
$${y_B} = 0.72$$
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