The correct option for the value of vapour pressure of a solution at $45^{0} C$ with benzene to octane in molar ratio 3 : 2 is
[At $45^{0} C$ , vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal gas.]

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

350 mm of Hg

336 mm of Hg

160 mm of Hg

168 mm of Hg

answer is D.

(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

Given

$n_{C_{6} H_{6}} {: n}_{C_{8} H_{8}} = 3 : 2$

$X_{Benzene} = \frac{3}{5} {;X}_{Octane} = \frac{2}{5}$

Vapour pressure of solution ${(P}_{s})$

${= P}_{C_{6} H_{6}}^{o} X_{C_{6} H_{6}} {+P}_{C_{8} H_{18}}^{o} X_{C_{8} H_{18}}$

$= 280 \times \frac{3}{5} + 420 \times \frac{2}{5}$

=168 + 168 = 336 mm of Hg

JEE

NEET

Foundation JEE

Foundation NEET

CBSE