Depression of freezing point of the solvent Questions in English

(N/A) The formula for molar mass $(M_2)$ using freezing point depression is:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given:
$K_f = 5.12 \, K \, kg \, mol^{-1}$
$w_2 = 1.00 \, g$
$w_1 = 50 \, g$
$\Delta T_f = 0.40 \, K$
Substituting the values:
$M_2 = \frac{5.12 \, K \, kg \, mol^{-1} \times 1.00 \, g \times 1000 \, g \, kg^{-1}}{0.40 \, K \times 50 \, g} = 256 \, g \, mol^{-1}$
Thus,the molar mass of the solute is $256 \, g \, mol^{-1}$.

(N/A) Mass of acetic acid,$w_{1} = 75 \ g$.
Molar mass of ascorbic acid $(C_{6}H_{8}O_{6})$,$M_{2} = (6 \times 12) + (8 \times 1) + (6 \times 16) = 176 \ g \ mol^{-1}$.
Lowering of melting point,$\Delta T_{f} = 1.5 \ K$.
Using the formula for depression in freezing point:
$\Delta T_{f} = \frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}$.
Rearranging for the mass of solute $(w_{2})$:
$w_{2} = \frac{\Delta T_{f} \times M_{2} \times w_{1}}{K_{f} \times 1000}$.
Substituting the values:
$w_{2} = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = \frac{19800}{3900} \approx 5.077 \ g$.
Hence,approximately $5.08 \ g$ of ascorbic acid is required.

(N/A) For cane sugar solution:
$\Delta T_{f} = (273.15 - 271) \, K = 2.15 \, K$.
$5 \%$ solution means $5 \, g$ of solute in $95 \, g$ of water.
Molar mass of cane sugar $(C_{12}H_{22}O_{11}) = 342 \, g \, mol^{-1}$.
Molality $(m) = \frac{5 / 342}{0.095} \, mol \, kg^{-1} = 0.1537 \, mol \, kg^{-1}$.
$K_{f} = \frac{\Delta T_{f}}{m} = \frac{2.15}{0.1537} = 13.99 \, K \, kg \, mol^{-1}$.
For glucose solution:
Molar mass of glucose $(C_{6}H_{12}O_{6}) = 180 \, g \, mol^{-1}$.
Molality $(m) = \frac{5 / 180}{0.095} \, mol \, kg^{-1} = 0.2926 \, mol \, kg^{-1}$.
$\Delta T_{f} = K_{f} \times m = 13.99 \times 0.2926 = 4.09 \, K$.
Freezing point of glucose solution $= 273.15 - 4.09 = 269.06 \, K$.

(N/A) The depression in freezing point $(\Delta T_f)$ is a colligative property,which depends on the number of particles (ions) produced in the solution.
Acetic acid $(CH_3COOH)$,trichloroacetic acid $(CCl_3COOH)$,and trifluoroacetic acid $(CF_3COOH)$ ionize to different extents.
The acidity depends on the electron-withdrawing effect of the substituent attached to the carboxylic group.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$,which is more electronegative than Hydrogen $(H)$.
Therefore,the electron-withdrawing inductive effect follows the order: $CF_3 > CCl_3 > CH_3$.
This makes the $O-H$ bond in trifluoroacetic acid the most polar,leading to the highest degree of ionization.
Since trifluoroacetic acid produces the maximum number of ions,it shows the greatest depression in freezing point.
Thus,the order of depression in freezing point is: $\text{Acetic acid} < \text{Trichloroacetic acid} < \text{Trifluoroacetic acid}$.

(N/A) The freezing point of a substance is 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.
When a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases.
Consequently,the solution freezes at a lower temperature than the pure solvent.
This difference in freezing point is known as the depression of freezing point,denoted by $\Delta T_f$.

(N/A) The molal depression constant $(K_f)$,also known as the cryoscopic constant,is defined as the depression in freezing point produced when one mole of a non-volatile solute is dissolved in one kilogram of a solvent.
Let $T_f^0$ be the freezing point of the pure solvent and $T_f$ be the freezing point of the solution. The depression in freezing point is given by $\Delta T_f = T_f^0 - T_f$.
For dilute solutions,the depression in freezing point is directly proportional to the molality $(m)$ of the solution:
$\Delta T_f \propto m$
$\Delta T_f = K_f \cdot m$ --- $(i)$
If $w_2$ grams of a solute with molar mass $M_2$ are dissolved in $w_1$ grams of solvent,the molality $(m)$ is:
$m = \frac{w_2 \times 1000}{M_2 \times w_1}$
Substituting this into equation $(i)$:
$\Delta T_f = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$
Rearranging to solve for $M_2$:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$

(B) $1$. Molar mass of glucose $(C_6H_{12}O_6) = (6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \ g \ mol^{-1}$.
$2$. Number of moles of glucose = $\frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
$3$. Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01 \ mol}{0.5 \ kg} = 0.02 \ m$.
$4$. Depression in freezing point $(\Delta T_f) = K_f \times m = 1.86 \ K \ kg \ mol^{-1} \times 0.02 \ mol \ kg^{-1} = 0.0372 \ K$.
$5$. Freezing point of solution $(T_f) = T_f^{\circ} - \Delta T_f = 273.15 \ K - 0.0372 \ K = 273.1128 \ K \approx 273.11 \ K$.

(A) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f$.
For sucrose ($C_{12}H_{22}O_{11}$,molar mass $M_1 = 342 \ g/mol$): $\Delta T_f = 273.15 \ K - 271 \ K = 2.15 \ K$.
Since $\Delta T_f = K_f \times m$,where $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,for a $5\% \ w/w$ solution,$m = \frac{5 \times 1000}{M_2 \times 95}$.
Thus,$\Delta T_f \propto \frac{1}{M_2}$.
For glucose ($C_6H_{12}O_6$,molar mass $M_2 = 180 \ g/mol$): $\frac{\Delta T_{f, \text{glucose}}}{\Delta T_{f, \text{sucrose}}} = \frac{M_{1}}{M_{2}} = \frac{342}{180} = 1.9$.
$\Delta T_{f, \text{glucose}} = 1.9 \times 2.15 \ K = 4.085 \ K$.
Therefore,$T_{f, \text{glucose}} = 273.15 \ K - 4.085 \ K = 269.065 \ K \approx 269.07 \ K$.

(72G) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f} = 0\,^\circ C - (-0.6\,^\circ C) = 0.6\,^\circ C$.
Using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality:
$0.6 = 1.5 \times \frac{w_{2} \times 1000}{M_{2} \times W_{1}(g)}$.
Substituting the values: $0.6 = 1.5 \times \frac{w_{2}}{60 \times 3}$.
$0.6 = 1.5 \times \frac{w_{2}}{180}$.
$w_{2} = \frac{0.6 \times 180}{1.5} = 0.4 \times 180 = 72\,g$.
Thus,$72\,g$ of urea is required.

(N/A) The relationship between the freezing point $(T_f)$ and the enthalpy of fusion $(\Delta_{fus}H)$ is given by the formula:
$T_f = \frac{\Delta_{fus}H}{\Delta_{fus}S}$
where $\Delta_{fus}S$ is the entropy of fusion.

(A) The depression in freezing point is given by the formula: $\Delta T_f = K_f \cdot m$,where $\Delta T_f$ is the depression in freezing point $(K)$,$K_f$ is the molal depression constant,and $m$ is the molality $(mol \cdot kg^{-1})$.
Rearranging for $K_f$: $K_f = \frac{\Delta T_f}{m}$.
Substituting the units: $K_f = \frac{K}{mol \cdot kg^{-1}} = K \cdot kg \cdot mol^{-1}$.

$1$. Calculate moles of acetic acid $(CH_3COOH)$: Molar mass $= 60 \ g \ mol^{-1}$. Mass $= 3 \times 10^{-3} \ kg = 3 \ g$. Moles $(n) = \frac{3 \ g}{60 \ g \ mol^{-1}} = 0.05 \ mol$.
$2$. Calculate mass of solvent (water): Volume $= 500 \ cm^3$,Density $= 0.997 \ g \ cm^{-3}$. Mass $= 500 \times 0.997 = 498.5 \ g = 0.4985 \ kg$.
$3$. Calculate molality $(m)$: $m = \frac{0.05 \ mol}{0.4985 \ kg} \approx 0.1003 \ mol \ kg^{-1}$.
$4$. Calculate Van't Hoff factor $(i)$: For $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$,$i = 1 \alpha = 1 0.23 = 1.23$.
$5$. Calculate depression in freezing point $(\Delta T_f)$: $\Delta T_f = i \times K_f \times m = 1.23 \times 1.86 \times 0.1003 \approx 0.229 \ K$.

(N/A) The phenomenon involved in clearing snow-covered roads is the $ \text{depression in freezing point} $ of water when a non-volatile solute is dissolved in it. When salt $( \text{NaCl} )$ is spread over snow-covered roads,it dissolves in the thin layer of water present on the surface of the ice. This increases the concentration of the solute,which lowers the freezing point of the water below the ambient temperature. Consequently,the ice melts even at temperatures below $ 0^{\circ}C $,helping to clear the roads.

(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$.
Given: $K_{f} = 5.12 \ K \ kg \ mol^{-1}$ and $m = 0.078 \ m$.
Substituting the values: $\Delta T_{f} = 5.12 \times 0.078$.
$\Delta T_{f} = 0.39936 \ K$.
Rounding off to two decimal places,we get $\Delta T_{f} = 0.40 \ K$.

(B) Given:
Mass of solute $(C_{4}H_{10})$ = $10 \ g$
Molar mass of $C_{4}H_{10}$ = $(4 \times 12) + (10 \times 1) = 58 \ g/mol$
Moles of solute = $\frac{10}{58} \ mol$
Mass of solvent $(C_{6}H_{6})$ = $200 \ g = 0.2 \ kg$
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{10/58}{0.2} = \frac{10}{58 \times 0.2} = \frac{10}{11.6} \approx 0.862 \ m$
Freezing point depression constant $(K_{f})$ = $5.12 \ ^{\circ} C/m$
Depression in freezing point $(\Delta T_{f})$ = $K_{f} \times m = 5.12 \times 0.862 \approx 4.41 \ ^{\circ} C$
Freezing point of solution $(T_{f})$ = $T_{f}^{\circ} - \Delta T_{f} = 5.5 - 4.41 = 1.09 \ ^{\circ} C$
Rounding to the nearest integer,the temperature is $1 \ ^{\circ} C$.

(B) $1$. Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$: $(2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g \ mol^{-1}$.
$2$. Calculate the number of moles of solute: $n = \frac{83 \ g}{62 \ g \ mol^{-1}} \approx 1.3387 \ mol$.
$3$. Calculate the molality $(m)$ of the solution: $m = \frac{n \text{ (mol)}}{W_{\text{solvent}} \text{ (kg)}} = \frac{1.3387 \ mol}{0.625 \ kg} = 2.1419 \ mol \ kg^{-1}$.
$4$. Calculate the depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f \times m = 1.86 \times 2.1419 \approx 3.984 \ K \approx 4 \ K$.
$5$. Calculate the freezing point of the solution $(T_f)$: $T_f = T_f^{\circ} - \Delta T_f = 273 \ K - 4 \ K = 269 \ K$.

(C) Initial molality $m = 0.75 \ mol \ kg^{-1}$.
Let the mass of water be $w_{1} \ kg$ and mass of sucrose be $w_{2} \ kg$.
Total mass $= w_{1} + w_{2} = 1 \ kg$.
Molality $m = \frac{w_{2} / 342}{w_{1}} = 0.75 \implies w_{2} = 0.75 \times 342 \times w_{1} = 256.5 \times w_{1}$.
Substituting into $w_{1} + 256.5 \times w_{1} = 1 \implies 257.5 \times w_{1} = 1 \implies w_{1} \approx 0.003883 \ kg$ (This interpretation is incorrect based on the problem statement).
Correct approach: $0.75 \ m$ solution means $0.75 \ mol$ sucrose in $1 \ kg$ water. Total mass $= 1000 + (0.75 \times 342) = 1256.5 \ g$.
Mass of sucrose in $1000 \ g$ solution $= \frac{0.75 \times 342}{1256.5} \times 1000 \approx 204.14 \ g$.
Mass of water in $1000 \ g$ solution $= 1000 - 204.14 = 795.86 \ g$.
Moles of sucrose $= 0.75 \times 0.79586 = 0.5969 \ mol$.
Freezing point depression $\Delta T_{f} = 4 = K_{f} \times m_{new} = 1.86 \times \frac{0.5969}{w_{new(kg)}}$.
$w_{new} = \frac{0.5969 \times 1.86}{4} = 0.2775 \ kg = 277.5 \ g$.
Ice separated $= 795.86 - 277.5 = 518.36 \ g \approx 518 \ g$.

(A) Step $1$: Calculate the number of moles of glucose.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \ g}{180 \ g \ mol^{-1}} = \frac{2}{9} \ mol \approx 0.222 \ mol$.
Step $2$: Calculate the mass of the solvent (water).
Density $= 1.00 \ g \ cm^{-3}$,so $200 \ mL$ of water $= 200 \ g = 0.2 \ kg$.
Step $3$: Calculate the molality $(m)$ of the solution.
$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}} = \frac{2/9 \ mol}{0.2 \ kg} = \frac{10}{9} \ m \approx 1.11 \ m$.
Step $4$: Calculate the depression in freezing point $(\Delta T_{f})$.
$\Delta T_{f} = K_{f} \times m = 1.86 \ K \ kg \ mol^{-1} \times \frac{10}{9} \ mol \ kg^{-1} \approx 2.067 \ K$.
Step $5$: Calculate the freezing point of the solution $(T_{f}^{\prime})$.
$T_{f}^{\prime} = T_{f} - \Delta T_{f} = 273.15 \ K - 2.067 \ K = 271.083 \ K$.
Rounding to the nearest integer,we get $271 \ K$.

(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.
Since the mass of solvent $(1 \ kg)$ and the mass of solute $(1 \ g)$ are the same for both,the molality $m$ is inversely proportional to the molar mass $M$ $(m = \frac{n}{W_{solvent}} = \frac{mass}{M \cdot W_{solvent}})$.
Therefore,$\frac{\Delta T_x}{\Delta T_y} = \frac{M_y}{M_x}$.
Given $\frac{\Delta T_x}{\Delta T_y} = \frac{1}{4}$,we have $\frac{1}{4} = \frac{M_y}{M_x}$.
This implies $\frac{M_x}{M_y} = \frac{4}{1}$,so $M_x : M_y = 4 : 1$.

(C) Step $1$: Calculate the moles of ascorbic acid $(C_{6}H_{8}O_{6})$.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{10.2 \ g}{176 \ g \ mol^{-1}} \approx 0.05795 \ mol$.
Step $2$: Calculate the molality $(m)$ of the solution.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05795 \ mol}{0.150 \ kg} \approx 0.3863 \ mol \ kg^{-1}$.
Step $3$: Calculate the depression in freezing point $(\Delta T_{f})$.
$\Delta T_{f} = K_{f} \times m = 3.9 \ K \ kg \ mol^{-1} \times 0.3863 \ mol \ kg^{-1} \approx 1.5066^{\circ} C$.
Step $4$: Express $\Delta T_{f}$ in the form $(x \times 10^{-1})^{\circ} C$.
$1.5066 \approx 15.066 \times 10^{-1}$.
Rounding to the nearest integer,$x = 15$.

(B) Mass of ethanol = $62.5 \ cm^{3} \times 0.80 \ g \ cm^{-3} = 50 \ g = 0.05 \ kg$.
Freezing point depression,$\Delta T_{f} = T_{f}^{\circ} - T_{f} = 156.0 \ K - 155.1 \ K = 0.9 \ K$.
Using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is molality:
$0.9 = 2.00 \times \frac{1.80 \ g / M_{w}}{0.05 \ kg}$.
$M_{w} = \frac{2.00 \times 1.80}{0.9 \times 0.05} = \frac{3.6}{0.045} = 80 \ g \ mol^{-1}$.

(D) The formula for molar mass $(M_2)$ using freezing point depression is: $M_2 = \frac{k_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given: $w_2 = 8.0 \, g$,$w_1 = 92 \, g$,$\Delta T_f = 0.925 \, ^{\circ}C$,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$.
Substituting the values: $M_2 = \frac{1.85 \times 8.0 \times 1000}{0.925 \times 92} = \frac{14800}{85.1} \approx 173.9 \, g \, mol^{-1}$.
Rounding to the nearest provided option,the molar mass is $160 \, g \, mol^{-1}$.

(A) Given:
Weight of solute $(w_2)$ = $0.643 \,g$
Freezing point constant $(K_f)$ = $5.12 \,K \,kg \,mol^{-1}$
Depression in freezing point $(\Delta T_f)$ = $5.51^{\circ}C - 5.03^{\circ}C = 0.48 \,K$
Volume of benzene = $50 \,mL$
Density of benzene = $0.879 \,g \,mL^{-1}$
Weight of solvent $(w_1)$ = $50 \,mL \times 0.879 \,g \,mL^{-1} = 43.95 \,g$
Using the formula for molar mass $(M_2)$:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
$M_2 = \frac{5.12 \times 0.643 \times 1000}{0.48 \times 43.95}$
$M_2 = \frac{3292.16}{21.096} \approx 156.05 \,g \,mol^{-1}$
Thus,the molar mass is approximately $156 \,g \,mol^{-1}$.

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.
Here,$i = 2.67$,$K_f = 1.8 \, K \, kg \, mol^{-1}$.
The molality $m$ is calculated as: $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in grams}}$.
For $38\%$ by weight $H_2SO_4$,mass of $H_2SO_4 = 38 \, g$,mass of water = $100 - 38 = 62 \, g$.
$m = \frac{38}{98} \times \frac{1000}{62} \approx 6.254 \, mol \, kg^{-1}$.
$\Delta T_f = 2.67 \times 1.8 \times 6.254 \approx 30.05 \, K$.
Freezing point of pure water = $273.15 \, K$.
Freezing point of solution = $273.15 - 30.05 = 243.1 \, K$.
Rounding to the nearest integer,we get $243 \, K$.

(D) When a non-volatile solute like naphthalene is added to a solvent like benzene,the freezing point of the solution becomes lower than that of the pure solvent. This phenomenon is known as the depression of freezing point.

(NONE) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Given: $\Delta T_{f} = 0 - (-24) = 24 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,Mass of solvent $(W_{solvent})$ $= 18.6 \ kg$,Molar mass of solute $(M_{solute})$ $= 62 \ g \ mol^{-1}$.
Molality $(m)$ $= \frac{W_{solute} \ (g)}{M_{solute} \times W_{solvent} \ (kg)} = \frac{W_{solute}}{62 \times 18.6}$.
Substituting the values: $24 = 1.86 \times \frac{W_{solute}}{62 \times 18.6}$.
$24 = \frac{1.86 \times W_{solute}}{1153.2}$.
$W_{solute} = \frac{24 \times 1153.2}{1.86} = 14880 \ g$.
Converting to $kg$: $W_{solute} = 14.88 \ kg$.

(A) Moles of water $= \frac{2700 \ g}{18 \ g \ mol^{-1}} = 150 \ mol$.
Moles of acetic acid $= \frac{2700 \ g}{60 \ g \ mol^{-1}} = 45 \ mol$.
Since the amount of water is greater than the amount of acetic acid,water acts as the solvent.
The depression in freezing point is given by $\Delta T_f = K_f \times m$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{45 \ mol}{2.7 \ kg} = 16.667 \ mol \ kg^{-1}$.
$\Delta T_f = 1.86 \ K \ kg \ mol^{-1} \times 16.667 \ mol \ kg^{-1} \approx 31 \ K$.
Freezing point of solution $= T_f^{\circ} - \Delta T_f = 0^{\circ} C - 31^{\circ} C = -31^{\circ} C$.
Therefore,$x = 31$.

(C) The freezing point depression is given by $\Delta T_f = T_f^\circ - T_f = 273.15 \ K - 270.65 \ K = 2.5 \ K$.
The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality.
$m = \frac{n_{methanol}}{w_{water} \text{ (in kg)}} = \frac{n}{0.1 \ kg}$.
Substituting the values: $2.5 = 1.86 \times \frac{n}{0.1} \Rightarrow n = \frac{2.5 \times 0.1}{1.86} \approx 0.1344 \ mol$.
The mass of methanol is $w = n \times M = 0.1344 \ mol \times 32 \ g \ mol^{-1} \approx 4.3008 \ g$.
The volume of methanol is $V = \frac{w}{d} = \frac{4.3008 \ g}{0.792 \ g \ cm^{-3}} \approx 5.4303 \ mL$.
Given $V = x \times 10^{-2} \ mL$,we have $5.4303 = x \times 10^{-2}$,so $x = 543.03$.
Rounding to the nearest integer,$x = 543$.

(C) The molality $(m)$ of the solution is calculated as:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{34.5 \ g / 46 \ g \ mol^{-1}}{0.5 \ kg} = \frac{0.75 \ mol}{0.5 \ kg} = 1.5 \ m$.
The depression in freezing point $(\Delta T_f)$ is given by:
$\Delta T_f = K_f \times m = 2 \ K \ kg \ mol^{-1} \times 1.5 \ mol \ kg^{-1} = 3 \ K$.
The new freezing point of the solution is $273 \ K - 3 \ K = 270 \ K$.
In the given plots,the freezing point is the temperature where the vapour pressure curve of the solution intersects the vapour pressure curve of ice. For a freezing point of $270 \ K$,the intersection must occur at $270 \ K$. Plot $(A)$ shows the intersection at $270 \ K$,and Plot $(C)$ also shows the intersection at $270 \ K$.

(D) The freezing point of a substance is the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
When a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases.
The freezing point of the solution $(T_f)$ is lower than the freezing point of the pure solvent $(T_f^\circ)$.
The graphical representation of depression in freezing point involves the intersection of the vapour pressure curve of the liquid solvent with the vapour pressure curve of the solid solvent (frozen solvent) to determine $T_f^\circ$,and the intersection of the vapour pressure curve of the solution with the vapour pressure curve of the solid solvent to determine $T_f$.
Option $D$ correctly depicts the curves for the liquid solvent,the solution,and the frozen solvent,showing the depression in freezing point $\Delta T_f = T_f^\circ - T_f$.

(A) The formula for freezing point depression is $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{1 / 256}{50 \times 10^{-3} \ kg} = \frac{1}{256 \times 0.05} = \frac{1}{12.8} \ mol \ kg^{-1} = 0.078125 \ mol \ kg^{-1}$.
Given $\Delta T_{f} = 0.40 \ K$.
Substituting the values: $0.40 = K_{f} \times 0.078125$.
$K_{f} = \frac{0.40}{0.078125} = 5.12 \ K \ kg \ mol^{-1}$.

(B) Statement $(I)$ is true because adding $NaCl$ to ice creates a freezing mixture,which lowers the temperature below $0^{\circ} C$,thereby keeping the ice cream frozen.
Statement $(II)$ is true because the addition of a non-volatile solute like $NaCl$ to a solvent like water causes a depression in the freezing point,a colligative property.

(D) Statement $-I$: The molal depression constant $K_f$ is defined as $K_f = \frac{M_1 R T_f^2}{\Delta H_{\text {fus }}}$. Since $\Delta S_{\text {fus }} = \frac{\Delta H_{\text {fus }}}{T_f}$,we can write $K_f = \frac{M_1 R T_f}{\Delta S_{\text {fus }}}$. Thus,Statement $-I$ is correct.
Statement $-II$: The value of $K_f$ for benzene is $5.12 \ \text{K kg mol}^{-1}$ and for water is $1.86 \ \text{K kg mol}^{-1}$. Since $5.12 > 1.86$,the $K_f$ for benzene is greater than that for water. Thus,Statement $-II$ is incorrect.

(C) The molar mass of ascorbic acid $(C_6H_8O_6)$ is $6 \times 12 + 8 \times 1 + 6 \times 16 = 176 \ g \ mol^{-1}$.
Using the formula for depression in freezing point: $\Delta T_f = K_f \times m$,where $m$ is the molality.
$m = \frac{w \times 1000}{M_{solute} \times W_{solvent(g)}}$,where $w$ is the mass of solute in grams.
Substituting the values: $1.5 = 3 \times \frac{w \times 1000}{176 \times 75}$.
$1.5 = \frac{3000 \times w}{13200}$.
$w = \frac{1.5 \times 13200}{3000} = \frac{19800}{3000} = 6.6 \ g$.

(A) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution after ice separates.
At $-0.5^{\circ}C$,$\Delta T_f = 0 - (-0.5) = 0.5 \ K$.
The mass of the remaining solvent $(x)$ is $200 \ g - 14 \ g = 186 \ g$.
Using the formula $\Delta T_f = K_f \times \frac{w \times 1000}{M \times x}$,where $w$ is the mass of glucose and $M = 180 \ g/mol$:
$0.5 = 1.86 \times \frac{w \times 1000}{180 \times 186}$.
$w = \frac{0.5 \times 180 \times 186}{1.86 \times 1000}$.
$w = \frac{0.5 \times 180 \times 186}{1860} = \frac{0.5 \times 180}{10} = 0.5 \times 18 = 9 \ g$.

(B) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f} = 7.734^{\circ} C - 5.43^{\circ} C = 2.304^{\circ} C$.
The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality.
Molality $m = \frac{w_{solute} \times 1000}{M_{solute} \times w_{solvent} \text{ (in g)}}$.
Substituting the values: $2.304 = 14.4 \times \frac{2.60 \times 1000}{M_{solute} \times 100}$.
$2.304 = 14.4 \times \frac{26}{M_{solute}}$.
$M_{solute} = \frac{14.4 \times 26}{2.304} = \frac{374.4}{2.304} = 162.5 \ g/mol$.

(C) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor and $m$ is the molality.
For $0.1 \ m \ NaCl$,$i = 2$,so $\Delta T_f = 2 \times 0.1 = 0.2$.
For $0.05 \ m \ MgSO_4$,$i = 2$,so $\Delta T_f = 2 \times 0.05 = 0.1$.
For $1 \ m \ AlPO_4$,$i = 2$,so $\Delta T_f = 2 \times 1 = 2.0$.
For $0.05 \ m \ Al_2(SO_4)_3$,$i = 5$,so $\Delta T_f = 5 \times 0.05 = 0.25$.
Comparing the values,$1 \ m \ AlPO_4$ has the highest value of $i \times m$,therefore it exhibits the highest freezing point depression.

(A) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given: $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}C - (-0.95^{\circ}C) = 0.95 \ K$.
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.95 = 1.86 \times m$.
$m = \frac{0.95}{1.86} \approx 0.51 \ mol \ kg^{-1}$.

(B) The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$, where $i$ is the van't Hoff factor and $m$ is the molality. Assuming complete dissociation, $i$ equals the number of ions produced per formula unit.
For $A$: $0.1 \, m \, NaClO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.1 \times 2 = 0.2$.
For $B$: $0.05 \, m \, MgSO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.05 \times 2 = 0.1$.
For $C$: $0.08 \, m \, AlPO_4 \rightarrow i = 2$, so $\Delta T_f \propto 0.08 \times 2 = 0.16$.
For $D$: $0.06 \, m \, Al_2(SO_4)_3 \rightarrow i = 5$, so $\Delta T_f \propto 0.06 \times 5 = 0.3$.
Comparing the values, $0.1$ is the lowest, which corresponds to option $B$.

(B) . The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \times m$,where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.
Given: $\Delta T_f = 1.8 \ K$ and $m = 0.4 \ m$.
Rearranging the formula to solve for $K_f$: $K_f = \frac{\Delta T_f}{m}$.
Substituting the values: $K_f = \frac{1.8 \ K}{0.4 \ mol \ kg^{-1}} = 4.5 \ K \ kg \ mol^{-1}$.
Therefore,the correct option is $B$.

(B) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_{solute} \times 1000}{M_{solute} \times W_{solvent(g)}}$.
Given: $\Delta T_{f} = 0.2 \ K$,$W_{solute} = 5 \ g$,$W_{solvent} = 50 \ g$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.2 = 1.86 \times \frac{5 \times 1000}{M_{solute} \times 50}$.
$0.2 = 1.86 \times \frac{100}{M_{solute}}$.
$M_{solute} = \frac{1.86 \times 100}{0.2} = \frac{186}{0.2} = 930 \ g \ mol^{-1}$.

(B) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} \times m$.
Here,$\Delta T_{f} = T_{f}^{\circ} - T_{f} = 0 \ ^{\circ}C - (-0.36 \ ^{\circ}C) = 0.36 \ K$.
Given $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.36 = 1.86 \times m$.
Therefore,$m = \frac{0.36}{1.86} \approx 0.1935 \ mol \ kg^{-1}$.
Rounding to three decimal places,the molality is $0.193 \ mol \ kg^{-1}$.

(C) The formula for depression in freezing point is given by: $\Delta T_f = K_f \times m$
Where:
$\Delta T_f$ is the depression in freezing point $(0.2 \ K)$
$K_f$ is the cryoscopic constant
$m$ is the molality of the solution $(0.18 \ m)$
Rearranging the formula to solve for $K_f$:
$K_f = \frac{\Delta T_f}{m} = \frac{0.2 \ K}{0.18 \ mol \ kg^{-1}} \approx 1.11 \ K \ kg \ mol^{-1}$
Therefore,the correct option is $C$.

(C) The formula for the depression in freezing point is $\Delta T_f = K_f \times m$.
Given:
Molality $(m)$ = $0.15 \ m$
Cryoscopic constant $(K_f)$ = $1.5 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_f = 1.5 \ K \ kg \ mol^{-1} \times 0.15 \ mol \ kg^{-1} = 0.225 \ K$.
Therefore,the correct option is $C$.

(C) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$.
Given: $\Delta T_{f} = 0.93 \ K$ and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Rearranging the formula for molality $(m)$:
$m = \frac{\Delta T_{f}}{K_{f}} = \frac{0.93 \ K}{1.86 \ K \ kg \ mol^{-1}} = 0.5 \ mol \ kg^{-1}$.

(B) The formula for molar mass of a solute is: $M_2 = \frac{1000 \times K_{f} \times W_2}{\Delta T_{f} \times W_1}$
Given: $W_2 = 1.5 \ g$,$W_1 = 90 \ g$,$\Delta T_{f} = 0.25 \ K$,$K_{f} = 1.2 \ K \ kg \ mol^{-1}$.
Substituting the values: $M_2 = \frac{1000 \times 1.2 \times 1.5}{0.25 \times 90}$
$M_2 = \frac{1800}{22.5} = 80 \ g \ mol^{-1}$.

(B) The formula for molar mass of a solute is given by: $M_2 = \frac{K_{f} \times W_2 \times 1000}{\Delta T_{f} \times W_1}$.
Given values are: $W_2 = 1 \ g$,$W_1 = 100 \ g$,$\Delta T_{f} = 0.2 \ K$,and $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.
Substituting these values into the formula:
$M_2 = \frac{1.2 \times 1 \times 1000}{0.2 \times 100} = \frac{1200}{20} = 60 \ g \ mol^{-1}$.

(B) The formula for the depression in freezing point is $\Delta T_{f} = i \times m \times K_{f}$.
Given:
van't Hoff factor $(i)$ = $1.1$
Molality $(m)$ = $0.01 \ m$
Cryoscopic constant $(K_{f})$ = $1.86 \ K \ kg \ mol^{-1}$
Substituting the values:
$\Delta T_{f} = 1.1 \times 0.01 \times 1.86$
$\Delta T_{f} = 0.02046 \ K \approx 0.020 \ K$.

(D) The formula for the depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality.
Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$,where $W_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent in grams.
Substituting the values: $\Delta T_f = 3 \ K$,$W_2 = 2.5 \ g$,$M_2 = 117 \ g \ mol^{-1}$,$W_1 = 35 \ g$.
$K_f = \frac{\Delta T_f \times M_2 \times W_1}{1000 \times W_2}$
$K_f = \frac{3 \times 117 \times 35}{1000 \times 2.5} \ K \ kg \ mol^{-1}$
$K_f = \frac{12285}{2500} \ K \ kg \ mol^{-1} = 4.91 \ K \ kg \ mol^{-1}$.

(D) The formula for depression in freezing point is $\Delta T_f = K_f \times m$.
Given: $\Delta T_f = 0.18 \ K$ and $K_f = 1.6 \ K \ kg \ mol^{-1}$.
Substituting the values in the formula:
$m = \frac{\Delta T_f}{K_f} = \frac{0.18 \ K}{1.6 \ K \ kg \ mol^{-1}} = 0.1125 \ m$.
Rounding to three decimal places,we get $m = 0.113 \ m$.