A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of…

Question

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

Accepted Answer

Correct answer: C. (c) Given: 1 mole of volatile liquid A 3 moles of volatile liquid B Vapor pressure of pure A, P_A^o=200 mmHg Vapor pressure of the solution, P_S=500 mm Hg We apply Raoult's law, which states: P_S=P_A^o X_A+P_B^o X_B Where: X_A is the mole fraction of A X_B is the mole fraction of B P_B^o is the vapor pressure of pure liquid B Calculate the mole fractions: X_A= 11+3= 14 X_B= 31+3= 34Plug these into the equation: 500=200 14+P_B^o 34Simplifying: 500=50+ 34P_B^oSubtract 50 from both sides: 450= 34P_B^oMultiply both sides by 43 to solve for P_B^o : P_B^o=600 mmHgSince P_A^o<P_B^o, liquid A is the least volatile component. In conclusion: The vapor pressure of pure B ,P_B^o, is 600 mm Hg The least volatile component is A.

$A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B . The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:$

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