The vapour pressure of Benzene at $25^{\circ}$ is 639.7mm of Hg. The vapour pressure of a solution of benzene at same temperature is 631.9 mm of Hg. The molality of solution is:

0.156 mol/kg

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1.56 mol/kg

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0.125 mol/kg

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None of these

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Solution

The correct option is A
0.156 mol/kg

$P_{1} = P_{01} x_{1} \Rightarrow \frac{P_{1}}{P_{01}} = x_{1} \Rightarrow x_{1} = \frac{631.9}{639.7} = 0.988 a n d x_{2} = 1 - x_{1} = 0.012 ∴ w e i g h t o f B e n z e n e = x_{1} \times 78 a n d m o l a l i t y = \frac{x_{2}}{x_{1} \times 78} \times 1000 = \frac{0.012}{78 \times 0.998} \times 1000 = 0.156 m o l / k g$

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The vapour pressure of Benzene at $25^{\circ}$ is 639.7mm of Hg. The vapour pressure of a solution of benzene at same temperature is 631.9 mm of Hg. The molality of solution is:

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The vapour pressure of Benzene at 25∘ is 639.7mm of Hg. The vapour pressure of a solution of benzene at same temperature is 631.9 mm of Hg. The molality of solution is:

The correct option is A 0.156 mol/kg P1=P01x1⇒P1P01=x1⇒x1=631.9639.7=0.988 and x2=1−x1=0.012∴weight of Benzene=x1×78and molality=x2x1×78×1000=0.01278×0.998×1000=0.156mol/kg

The vapour pressure of Benzene at $25^{\circ}$ is 639.7mm of Hg. The vapour pressure of a solution of benzene at same temperature is 631.9 mm of Hg. The molality of solution is: