Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

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

$160 \text{ mm of Hg}$

$168 \text{ mm of Hg}$

$336 \text{ mm of Hg}$

$350 \text{ mm of Hg}$

Step-by-Step Solution

Given $n_{\text{C}_6\text{H}_6} : n_{\text{C}_8\text{H}_{18}} = 3:2$ . Mole fractions: $x_{\text{C}_6\text{H}_6} = 3/5$ , $x_{\text{C}_8\text{H}_{18}} = 2/5$ . Total pressure $P_s = P^{\circ}_A x_A + P^{\circ}_B x_B = 280 \times (3/5) + 420 \times (2/5) = 168 + 168 = 336 \text{ mm Hg}$ .

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