A solution is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$ . Given vapour pressure of pure water and vapour pressure of pure ethanol are $32.8$ mm $Hg$ and $40 mm Hg$ respectively. The vapour pressure of solution M 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

$29.5 mm Hg$

$39.3 mm Hg$

$36.0 mm Hg$

$28.8 mm Hg$

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

P = total vapour pressure.

$P = p_{A}^{°} X_{A} + p_{B}^{°} X_{B} = 40 \times 0.9 + 32.8 \times 0.1 (since X = X_{A} + X_{B}) = 39.28 mm Hg ≃ 39.3 m m H g$

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