Answer:

the freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.

Let ${\mathrm{T}}_{\mathrm{f}}^0$ be the freezing point of pure solvent and T_f be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point.
$∆{\mathrm{T}}_{\mathrm{f}} = {\mathrm{T}}_{\mathrm{f}}^0 - {\mathrm{T}}_{\mathrm{f}}$

Similar to elevation of boiling point, depression of freezing point (ΔT_f) for dilute solution (ideal solution) is directly proportional to molality,
m of the solution. Thus,
ΔT_f ∝ m

or
ΔT_f= K_fm

The proportionality constant, K_f, which depends on the nature of the solvent is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant. The unit of Kf is K kg mol^-1

If w₂ gram of the solute having molar mass as M₂, present in w₁ gram of solvent, produces the depression in freezing point ΔT_f of the
solvent then molality of the solute is given by the equation:
$$\begin{gathered} \mathrm{m}= \frac{{\displaystyle \raisebox{1ex}{${\mathrm{w}}_2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{M}}_2$}\right.}}{{\displaystyle \raisebox{1ex}{${\mathrm{w}}_1$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}} \\ \mathrm{substituting} \mathrm{this} \mathrm{value} \mathrm{in} ∆{\mathrm{T}}_{\mathrm{f}} = {\mathrm{K}}_{\mathrm{f}}\mathrm{m} \\ ∆{\mathrm{T}}_{\mathrm{f}} = \frac{{\mathrm{K}}_{\mathrm{f}} \times {\mathrm{w}}_2 \times 1000}{{\mathrm{M}}_2\times {\mathrm{w}}_1} \\ {\mathrm{M}}_2 = \frac{{\mathrm{K}}_{\mathrm{f}} \times {\mathrm{w}}_2 \times 1000}{∆{\mathrm{T}}_{\mathrm{f}}\times {\mathrm{w}}_1} \end{gathered}$$

Thus for determining the molar mass of the solute we should know the quantities w₁, w₂, ΔT_f, along with the molal freezing point depression constant.

A.

D.

D.

Answer:

Raoult's law state that for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

p₁ ∝ x₁

and p₁ = ${\mathrm{p}}_1^0$x₁

The proportionality constant is equal to the vapour pressure of pure solvent, ${\mathrm{p}}_1^0$.
IDEAL SOLUTION

(i) The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

Δ_mixH = 0
Δ_mixV = 0

It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of
volumes of the two components.

(ii) 

At molecular level, ideal behaviour of
the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will
be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of ideal
solution.
example: Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.

NON-IDEAL SOLUTION

When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-ideal solution.
In non-ideal solution the vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law, If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation.

IN case of non-ideal solution

Δ_mixH ,Δ_mixV
 both are not equal to zero. it either be less than zero or greater than zero.
 IN case of positive deviation it be larger and In case of negtive deviation it be lesser.
in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.
If the intermolecular attractive forces between the A-A and B-B are not nearly equal to those between A-B, this leads to the formation of non ideal solution.

example :a mixture of chloroform and acetone etc.