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

Question

The correct option for the value of vapour pressure of a solution at 45° C with benzene to octane in molar ratio 3: 2 is: [At 45° C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas] (A) 160 mm of Hg (B) 168 mm of Hg (C) 336 mm of Hg (D) 350 mm of Hg

Tardigrade Admin · Accepted Answer

Applying Raoult's law, P text total =PA0 . XA+PB0 . XB XA=(3/5), XB=(2/5) P text total =280 ×((3/5))+420 ×((2/5)) =336 mm of Hg

Q. The correct option for the value of vapour pressure of a solution at $4 5^{\circ} C$ with benzene to octane in molar ratio $3 : 2$ is : [At $4 5^{\circ} C$ vapour pressure of benzene is $280 mm H g$ and that of octane is $420 mm H g$ . Assume Ideal gas]

160 mm of Hg

168 mm of Hg

336 mm of Hg

350 mm of Hg

Applying Raoult's law,
$P_{total} = P_{A}^{0} . X_{A} + P_{B}^{0} . X_{B}$
$X_{A} = \frac{3}{5}, X_{B} = \frac{2}{5}$
$P_{total} = 280 \times (\frac{3}{5}) + 420 \times (\frac{2}{5})$
$= 336 mm$ of $H g$

Q. The correct option for the value of vapour pressure of a solution at 45∘C with benzene to octane in molar ratio 3:2 is : [At 45∘C vapour pressure of benzene is 280mmHg and that of octane is 420mmHg. Assume Ideal gas]