Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of pure A and pure B are 7×10³ Pa and 12 × 10³ Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

x_A= 0.76 ; x_B= 0.24

x_A= 0.37 ; x_B= 0.63

x_A= 0.28 ; x_B= 0.72

x_A= 0.4 ; x_B= 0.6

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Complete Solution

Given Data:

x_A = 0.4
P_A⁰ = 7 × 10³ Pa
P_B⁰ = 12 × 10³ Pa

Total Pressure:

P = (0.4 × 7 × 10³) + (0.6 × 12 × 10³) = 10 × 10³ Pa

Mole Fraction in Vapor Phase:

x_A = (0.4 × 7 × 10³) / (10 × 10³) = 0.28

x_B = 1 - 0.28 = 0.72

Hence, correct option is: (B) X_A = 0.28 and X_B = 0.72

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!