5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$134 g {mol}^{- 1}$

$130 g {mol}^{- 1}$

$120 g {mol}^{- 1}$

$125 g {mol}^{- 1}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Mole fraction of water in solution is

$x = \frac{(100 / 18) mol}{(5 g / M) + (100 / 18) mol}$

Hence, $p = x_{1} p_{1}^{*}$

$2980 Pa = \frac{(100 / 18) mol}{(5 g / M) + (100 / 18) mol} (3000 Pa)$ This gives $M = 134.1 g {mol}^{- 1}$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test