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A binary mixture of acetone and chloroform is found to exhibit negative deviation from Raoult's Law. The mole fraction of acetone in the mixture is 0.6, and the vapour pressure of pure acetone and pure chloroform are 350 mmHg and 200 mmHg, respectively. What is the vapour pressure of the mixture?

Option: 1

275 \mathrm{~mm}$ of $\mathrm{Hg}

Option: 2

300 \mathrm{~mm}$ of $\mathrm{Hg}

Option: 3

225 \mathrm{~mm}$ of $\mathrm{Hg}

Option: 4

290 \mathrm{~mm}$ of $\mathrm{Hg}

Answers (1)


The vapour pressure of the mixture can be calculated using the equation: P_{\text {total }}=x_A \times P_A^{\gamma_A}+x_B \times P_B^{\gamma_B}, where x_A \text { and } x_Bare the mole fractions of the components, P_A^{\gamma_A}and  P_B^{\gamma_B}are the vapour pressures of the pure components, and \gamma_A and  \gamma_B are the activity coefficients of the components. In this case, since the mixture exhibits negative deviation from Raoult's Law, the activity coefficients are less than 1. Assuming the activity coefficients are equal, we can solve for P_{\text {total }} as follows:

P_{\text {total }}=(0.6 \times 350)+(0.4 \times 200 )= 210+80=290 \mathrm{~mm} \text { of } \mathrm{Hg}

Therefore, the vapour pressure of the mixture is less than the vapour pressure predicted by Raoult's Law, indicating negative deviation.


Posted by

Irshad Anwar

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