At 300 K,the vapour pressure of a pure liqui | vedclass

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Question

(A) According to Raoult's Law for an ideal solution,the total vapour pressure $P_{total}$ is given by: $P_{total} = P_A^0 \cdot x_A + P_B^0 \cdot x_B$

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(A) According to Raoult's Law for an ideal solution,the total vapour pressure $P_{total}$ is given by: $P_{total} = P_A^0 \cdot x_A + P_B^0 \cdot x_B$ Given: $P_A^0 = 70 \ mm \ Hg$ $x_B = 0.2$ $x_A = 1 - x_B = 1 - 0.2 = 0.8$ $P_{total} = 84 \ mm \ Hg$ Substituting the values: $84 = (70 	imes 0.8) + (P_B^0 	imes 0.2)$ $84 = 56 + 0.2 \cdot P_B^0$ $84 - 56 = 0.2 \cdot P_B^0$ $28 = 0.2 \cdot P_B^0$ $P_B^0 = \frac{28}{0.2} = 140 \ mm \ Hg$ Thus,the vapour pressure of pure liquid $B$ is $140 \ mm \ Hg$.

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