Solutions

When

and

from an ideal solution,

K2Hgl4 is 40% ionised. $\therefore$ $\alpha$ =

= 0.4 K2[Hgl4] $\to$ 2K+ + [Hgl4]2+ N =

= 3 i = 1 + (N - 1)$\alpha$ = 1 + (3 - 1)0.4 = 1 + 2$\times$0.4 = 1.8

$\therefore$

For a binary solution of two volatile liquid components labeled 1 and 2, let $x_1$ and $y_1$ represent the mole fractions of component 1 in the liquid and vapor phases, respectively.

The linear relationship between the inverse of these mole fractions is plotted as $\dfrac{1}{x_1}$ versus $\dfrac{1}{y_1}$.

To derive the slope and intercept of this linear plot, consider the following calculations: Using Raoult's Law for a Liquid Solution: For a liquid solution with volatile components 1 and 2: $\mathrm{P}_1 = \mathrm{P}_{\mathrm{T}} \cdot y_1 = \mathrm{P}_1^{\mathrm{o}} \cdot x_1$ Therefore: $\dfrac{\mathrm{P}_{\mathrm{T}}}{x_1} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Rearranging the Equation: By substituting and rearranging, we have: $\dfrac{\mathrm{P}_2^{\mathrm{o}} + x_1(\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}})}{x_1} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Simplifying further: $\dfrac{\mathrm{P}_2^{\mathrm{o}}}{x_1} + (\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}}) = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Expressing $\dfrac{1}{x_1}$: Solving for $\dfrac{1}{x_1}$, we obtain: $\dfrac{1}{x_1} = \left(\dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)\left(\dfrac{1}{y_1}\right) + \left(\dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)$ Determining the Slope and Intercept: The slope of the line is: $\text{Slope} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ The intercept of the line is: $\text{Intercept} = \dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ In summary, for the plot of $\dfrac{1}{x_1}$ against $\dfrac{1}{y_1}$, the slope is $\dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ and the intercept is $\dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$.

Moles of glucose

Moles of water

Total moles

Mole fraction $\times$ Total pressure

Torr

Given, Vapour pressure of benzene

Vapour pressure of toluene

mass of benzene in

hence moles of benzene

(mol. wt of benzene

) mass of toluene in solution

hence moles of toluene

Total moles of benzene and toluene = 1.5 mol now partial pressure of benzene

torr

In the context of solutions and mixtures, a negative deviation from Raoult's law occurs when the interaction between different molecules (A-B interaction) is stronger than the interaction between similar molecules (A-A and B-B interactions).

This is because the molecules prefer to interact with each other rather than themselves, leading to a decrease in the overall vapor pressure compared to what would be expected from Raoult's law.

So, the correct option is : Option D : A - B interaction is stronger than A - A and B - B interaction Option A suggests that the change in volume ($\Delta V_{mix}$) and change in entropy ($\Delta S_{mix}$) of the mixture are both greater than zero.

This doesn't necessarily indicate a negative deviation from Raoult's law.

It's possible for these conditions to be true in both positive and negative deviations or even in ideal solutions.

Therefore, this option isn't specifically associated with negative deviations.

Option B suggests that there's no change in the volume of the mixture ($\Delta V_{mix} = 0$) and the change in entropy ($\Delta S_{mix}$) is greater than zero.

Again, this doesn't specifically indicate a negative deviation from Raoult's law.

The volume change upon mixing can be zero, positive, or negative depending on the nature of the interactions among the molecules.

Raoult's law deviations are primarily dictated by the relative strength of the intermolecular forces among the different components in the mixture.

Thus, the change in volume and entropy, while they can be related, do not directly determine whether a solution will show a positive or negative deviation.

NOTE : On increasing pressure, the temperature is also increased.

Thus in pressure cooker due to increase in pressure the

of water increases.

Since

has different values for different solvents, hence even if the

is the same

will be different

For this solution intermolecular interactions between

-heptane and ethanol aare weaker than

-heptane -

- heptane & ethanol-ethanol interactions hence the solution of

-heptane and ethanol is non-ideal and shows positive deviation from Raoult's law.