Which one of the following graphs accurately represents the plot of partial pressure of $$CS_{2}$$ vs its mole fraction in a mixture of acetone and $$CS_{2}$$ at constant temperature?

For any binary liquid mixture maintained at a fixed temperature, Raoult’s law for an ideal solution states that the partial pressure of a component equals the product of its mole fraction and its pure-component vapour pressure:

$$p_i = x_i \, p_i^{\,*} \qquad (i = 1,2)$$

Thus, for an ideal solution of $$CS_2$$ (component-2) with any other liquid, the graph of $$p_{CS_2}$$ versus $$x_{CS_2}$$ would be a straight line passing through the origin with slope $$p_{CS_2}^{\,*}$$.

The mixture in the question, however, is acetone + $$CS_2$$. Acetone molecules are strongly dipolar, while $$CS_2$$ molecules are non-polar. The unlike (acetone-$$CS_2$$) interactions are therefore much weaker than the like (acetone-acetone dipole-dipole and $$CS_2$$-$$CS_2$$ dispersion) interactions. Because of these weaker intermolecular forces, molecules escape more readily into the vapour phase, raising the vapour pressure above the Raoult-law value.

This behaviour is called a positive deviation from Raoult’s law. Mathematically,

$$p_{CS_2} \; \gt \; x_{CS_2}\,p_{CS_2}^{\,*}$$

for every intermediate value of $$x_{CS_2}$$ (except at the ends, where the mixture is pure $$CS_2$$ and equality holds).

Consequently, the plot of $$p_{CS_2}$$ versus $$x_{CS_2}$$ must lie above the straight-line ideal plot and be curved upwards (convex to the mole-fraction axis).

Among the four sketches given in the question, the only one that shows an upward-curving line starting at the origin, lying above the ideal straight line, and reaching $$p_{CS_2}^{\,*}$$ at $$x_{CS_2}=1$$ is Option C.

Hence, the correct choice is:
Option C which is: (graph showing positive deviation from Raoult’s law).