Depression of freezing point of the solvent Questions in English

(D) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 5.5 ^\circ C - 4.48 ^\circ C = 1.02 ^\circ C$ (or $1.02 \ K$).
Using the formula for molar mass $M = \frac{K_f \times w \times 1000}{W \times \Delta T_f}$,where $w = 4.8 \ g$,$W = 60 \ g$,$K_f = 5.1 \ K \ kg \ mol^{-1}$,and $\Delta T_f = 1.02 \ K$.
$M = \frac{5.1 \times 4.8 \times 1000}{60 \times 1.02} = \frac{24480}{61.2} = 400 \ g \ mol^{-1}$.

(D) The depression in freezing point is given by the formula $\Delta T_f = m \times K_f$,where $m$ is the molality of the solution and $K_f$ is the cryoscopic constant of the solvent.
For the first case: $m_1 = \frac{0.01 \ mol}{100 \ g} \times 1000 \ g/kg = 0.1 \ mol/kg$.
Given $\Delta T_{f1} = 0.40 \ ^oC$,we have $0.40 = 0.1 \times K_f$,which gives $K_f = 4 \ ^oC \ kg/mol$.
For the second case: $m_2 = \frac{0.03 \ mol}{50 \ g} \times 1000 \ g/kg = 0.6 \ mol/kg$.
Now,calculating the new depression in freezing point: $\Delta T_{f2} = m_2 \times K_f = 0.6 \times 4 = 2.4 \ ^oC$.

(C) The molal depression constant,also known as the cryoscopic constant $(K_f)$,is a characteristic property of the solvent.
It depends solely on the nature of the solvent and is independent of the concentration of the solute or the molality of the solution.
Therefore,if the molality of the dilute solution is doubled,the value of $(K_f)$ will remain unchanged.

(A) Given: $w_2 = 68.5 \, g$,$M_2 = 342 \, g \, mol^{-1}$,$w_1 = 1000 \, g = 1 \, kg$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.
$\Delta T_f = K_f \times m = K_f \times \frac{w_2}{M_2 \times w_1(kg)}$
$\Delta T_f = 1.86 \times \frac{68.5}{342 \times 1} = 0.3725 \, K$ or $^oC$.
Freezing point of solution $T_f = T_f^o - \Delta T_f = 0 \, ^oC - 0.3725 \, ^oC = -0.3725 \, ^oC \approx -0.37 \, ^oC$.

(B) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_1 \times 1000}{M_1 \times W_2}$,where $W_1$ is the mass of solute,$M_1$ is the molar mass of solute,and $W_2$ is the mass of solvent in grams.
Given: $W_1 = 1.00 \ g$,$M_1 = 250 \ g \ mol^{-1}$,$W_2 = 51.2 \ g$,$K_f = 5.12 \ K \ kg \ mol^{-1}$.
Substituting the values: $\Delta T_f = 5.12 \times \frac{1.00 \times 1000}{250 \times 51.2}$.
$\Delta T_f = 5.12 \times \frac{1000}{12800} = 5.12 \times 0.078125 = 0.4 \ K$.

(A) Given:
$K_f = 1.86 \ K \ kg \ mol^{-1}$
Mass of solvent $(W_A)$ $= 4 \ kg = 4000 \ g$
Freezing point depression $\Delta T_f = 0 - (-6) = 6 \ K$
Molar mass of solute $(M_B)$ $= 62 \ g \ mol^{-1}$
The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is molality.
$\Delta T_f = K_f \times \frac{w \times 1000}{M_B \times W_A \text{ (in g)}}$
$6 = 1.86 \times \frac{w \times 1000}{62 \times 4000}$
$6 = 1.86 \times \frac{w}{62 \times 4}$
$w = \frac{6 \times 62 \times 4}{1.86}$
$w = \frac{1488}{1.86} = 800 \ g$

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given values: $\Delta T_f = 2.8 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$,$i = 1$ (since ethylene glycol is a non-electrolyte).
Mass of solvent (water) $= 1.0 \, kg$.
Let the mass of solute (ethylene glycol) be $x \, g$.
Molar mass of ethylene glycol $(C_2H_6O_2)$ $= (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \, g \, mol^{-1}$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{x / 62}{1} = \frac{x}{62} \, mol \, kg^{-1}$.
Substituting into the equation: $2.8 = 1 \times 1.86 \times \frac{x}{62}$.
Solving for $x$: $x = \frac{2.8 \times 62}{1.86} = 93.33 \, g \approx 93 \, g$.

(C) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$. The freezing point is $T_f = T_f^0 - \Delta T_f$. To have the highest freezing point,the value of $\Delta T_f$ must be the lowest,which means the van't Hoff factor $(i)$ multiplied by molality $(m)$ must be the minimum.
For $A$: $i = 3$ $(Ca^{2+} + 2NO_3^-)$,$i \times m = 3 \times 0.12 = 0.36$.
For $B$: $i = 2$ $(Na^+ + Cl^-)$,$i \times m = 2 \times 0.15 = 0.30$.
For $C$: $i = 1$ (urea is non-electrolyte),$i \times m = 1 \times 0.2 = 0.20$.
For $D$: $i \approx 1$ (weak acid,$CH_3COOH$ is slightly dissociated),$i \times m \approx 1 \times 0.2 = 0.20$. Since $CH_3COOH$ dissociates slightly,$i > 1$,so $i \times m > 0.20$.
Comparing the values,urea has the lowest $i \times m$ value,thus it has the highest freezing point.

(C) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given $\Delta T_f = 0.5 \ ^\circ C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and moles of glucose = $\frac{36 \ g}{180 \ g/mol} = 0.2 \ mol$.
Let $W$ be the mass of water remaining in liquid state (in $kg$).
$0.5 = 1.86 \times \frac{0.2}{W}$
$W = \frac{1.86 \times 0.2}{0.5} = 0.744 \ kg = 744 \ g$.
Since the initial mass of water was $1000 \ g$,the mass of ice separated = $1000 \ g - 744 \ g = 256 \ g$.

(B) We know that $\Delta T_f = K_f \cdot m$. Since the amount of solute is constant, $\Delta T_f \propto \frac{1}{W}$, where $W$ is the mass of the solvent. Given: $\Delta T_{f1} = 0.2^\circ\text{C}$, $W_1 = 100 \text{ g}$, $\Delta T_{f2} = 0.25^\circ\text{C}$. $\frac{\Delta T_{f1}}{\Delta T_{f2}} = \frac{W_2}{W_1}$ $\frac{0.2}{0.25} = \frac{W_2}{100}$ $W_2 = \frac{0.2 \cdot 100}{0.25} = 80 \text{ g}$ Mass of ice separated = $W_1 - W_2 = 100 \text{ g} - 80 \text{ g} = 20 \text{ g}$.

(B) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$.
Freezing point $T_f = T_f^0 - \Delta T_f = 273 - (i \times K_f \times 0.1)$.
For sucrose,$i=1$,$\Delta T_f = 273 - 272 = 1 \, K$.
Thus,$K_f \times 0.1 = 1$.
$T_f = 273 - i$.
For $a. \ BaCl_2$ $(i=3)$: $T_f = 273 - 3 = 270 \, K$ $(q)$.
For $b. \ NaCl$ $(i=2)$: $T_f = 273 - 2 = 271 \, K$ $(p)$.
For $c. \ K_3[Fe(CN)_6]$ $(i=4)$: $T_f = 273 - 4 = 269 \, K$ $(s)$.
For $d. \ Al_2(SO_4)_3$ $(i=5)$: $T_f = 273 - 5 = 268 \, K$ $(r)$.
Therefore,the correct match is $a-q, b-p, c-s, d-r$.

(C) The provided diagram represents the vapour pressure curve for the depression in freezing point.
In this graph:
$(i)$ represents the intersection point of the vapour pressure curve of the solid solvent and the solution,which corresponds to the freezing point of the solution $(T_f)$.
$(ii)$ represents the vapour pressure curve of the pure liquid solvent.
$(iii)$ represents the vapour pressure curve of the solution.
Therefore,$(i)$ is the frozen solvent (at the freezing point),$(ii)$ is the liquid solvent,and $(iii)$ is the solution.
Thus,the correct identification is $(i)$ Frozen solvent,$(ii)$ Liquid solvent,$(iii)$ Solution.

(A) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 271.14 \ K = 1.86 \ K$.
Using the formula $\Delta T_f = K_f \times m$,where $m$ is the molality:
$1.86 = 1.86 \times m \implies m = 1 \ mol/kg$.
This means $1 \ mol$ of urea is dissolved in $1000 \ g$ of water.
Moles of water = $\frac{1000 \ g}{18 \ g/mol} \approx 55.55 \ mol$.
Mole fraction of urea $(X_{\text{urea}})$ = $\frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx \frac{1}{56.5}$.

(B) The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.
Given: $\Delta T_{f} = 0 - (-0.93) = 0.93\,^{\circ}C$,$K_{f} = 1.86\,^{\circ}C\,kg\,mol^{-1}$,and the solution is $7.0\%$ by mass,meaning $7\,g$ of solute in $93\,g$ of water.
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{7}{M \times 0.093}$.
Substituting the values: $0.93 = 1.86 \times \frac{7}{M \times 0.093}$.
$M = \frac{1.86 \times 7}{0.93 \times 0.093} = \frac{13.02}{0.08649} \approx 150.5\,g\,mol^{-1}$.

(A) The freezing point depression is given by $\Delta T_f = i \times K_f \times m$. The freezing point of the solution is $T_f = T_f^\circ - \Delta T_f$. To have the minimum freezing point,the solution must have the maximum value of the van't Hoff factor $(i)$ multiplied by molality $(m)$.
$A: 0.01 \ M \ NaCl \implies i=2, m=0.01, i \times m = 0.02$
$B: 0.005 \ M \ C_2H_5OH \implies i=1, m=0.005, i \times m = 0.005$
$C: 0.005 \ M \ MgI_2 \implies i=3, m=0.005, i \times m = 0.015$
$D: 0.005 \ M \ MgSO_4 \implies i=2, m=0.005, i \times m = 0.010$
Comparing the values of $i \times m$,$0.01 \ M \ NaCl$ has the highest value,therefore it will have the maximum depression in freezing point and the minimum freezing point.

(D) The molar mass of fructose $(C_6H_{12}O_6)$ is $180 \ g \ mol^{-1}$.
Number of moles of fructose = $\frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01 \ mol}{2 \ kg} = 0.005 \ m$.
Depression in freezing point $(\Delta T_f)$ = $i \times m \times k_f$.
Since fructose is a non-electrolyte,the van't Hoff factor $(i)$ = $1$.
$\Delta T_f = 1 \times 0.005 \times 1.86 = 0.0093 \ K$ (or $^\circ C$).
Freezing point of solution = $T_f^\circ - \Delta T_f = 0 \ ^\circ C - 0.0093 \ ^\circ C = -0.0093 \ ^\circ C$.

(D) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given: $\Delta T_f = 273 \, K - 268 \, K = 5 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$,volume of water = $10 \, L$,mass of water $(W)$ = $10 \, kg$,molar mass of methyl alcohol $(CH_3OH)$ = $32 \, g/mol$.
Molality $m = \frac{w \times 1000}{M \times W_{solvent(g)}} = \frac{w}{32 \times 10}$.
Substituting the values: $5 = 1.86 \times \frac{w}{32 \times 10}$.
$w = \frac{5 \times 32 \times 10}{1.86} = \frac{1600}{1.86} \approx 860.22 \, g$.
Comparing with the given options,the closest value is $868.06 \, g$.

(C) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62 \ g \ mol^{-1}$.
Moles of ethylene glycol $= \frac{62 \ g}{62 \ g \ mol^{-1}} = 1 \ mol$.
Let the mass of water remaining in the liquid phase be $w \ g$.
The freezing point depression is $\Delta T_f = 0 - (-10) = 10 \ K$.
Using the formula $\Delta T_f = K_f \times m$,where $m$ is the molality:
$10 = 1.86 \times \frac{1 \ mol}{(w / 1000) \ kg}$.
$w = \frac{1.86 \times 1000}{10} = 186 \ g$.
The amount of water separated as ice $= \text{Initial mass} - \text{Remaining mass} = 250 \ g - 186 \ g = 64 \ g$.

(B) The depression in freezing point $\Delta T_f$ is directly proportional to the concentration of the solute in the milk. Let $V_p$ be the volume of pure milk and $V_w$ be the volume of added water.
For pure milk,the freezing point depression is proportional to the concentration $C_p = \frac{n}{V_p} = 0.5$.
For diluted milk,the concentration is $C_d = \frac{n}{V_p + V_w} = 0.2$.
Taking the ratio: $\frac{C_p}{C_d} = \frac{V_p + V_w}{V_p} = \frac{0.5}{0.2} = 2.5$.
This implies $V_p + V_w = 2.5 V_p$,so $V_w = 1.5 V_p$.
If $V_p = 2$ cups,then $V_w = 1.5 \times 2 = 3$ cups.
Thus,$3$ cups of water are added to $2$ cups of pure milk.

(B) The freezing point of a substance is related to its ability to form a stable crystal lattice,which is influenced by intermolecular forces and molecular symmetry.
Phthalic acid $(C_8H_6O_4)$ has strong intermolecular hydrogen bonding.
$2$-Naphthol $(C_{10}H_8O)$ also exhibits hydrogen bonding.
Naphthalene $(C_{10}H_8)$ is a highly symmetric planar molecule that packs well in a crystal lattice.
$9$,$10$-Dimethylanthracene is a bulky,non-polar molecule with significant steric hindrance due to the methyl groups at the $9$ and $10$ positions,which disrupts efficient crystal packing,leading to a lower freezing point compared to the others.

(A) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Since the freezing points are equal,$(\Delta T_f)_X = (\Delta T_f)_Y$,which implies $m_X = m_Y$.
For a $4 \%$ aqueous solution of $X$,the mass of $X$ is $4 \ g$ in $96 \ g$ of water. Molality $m_X = \frac{4 \times 1000}{A \times 96}$.
For a $12 \%$ aqueous solution of $Y$,the mass of $Y$ is $12 \ g$ in $88 \ g$ of water. Molality $m_Y = \frac{12 \times 1000}{M_Y \times 88}$.
Equating the two: $\frac{4}{A \times 96} = \frac{12}{M_Y \times 88}$.
Solving for $M_Y$: $M_Y = \frac{12 \times 96 \times A}{4 \times 88} = \frac{3 \times 96 \times A}{88} = \frac{288 \times A}{88} \approx 3.27 \times A$.
Rounding to the nearest integer,the molecular weight of $Y$ is $3A$.

(C) The freezing point depression is given by $\Delta T_f = i \times K_f \times m$. The freezing point is $T_f = T_f^0 - \Delta T_f$. Thus, a higher value of the van't Hoff factor $(i)$ multiplied by molality $(m)$ results in a lower freezing point.
Calculate the product $i \times m$ for each solution:
$(a) \ CuSO_4 \rightarrow Cu^{2+} + SO_4^{2-}$, so $i = 2$. Product $= 2 \times 0.075 = 0.150 \ M$.
$(b) \ (NH_4)_2SO_4 \rightarrow 2NH_4^+ + SO_4^{2-}$, so $i = 3$. Product $= 3 \times 0.060 = 0.180 \ M$.
$(c) \ \text{Urea is a non-electrolyte}$, so $i = 1$. Product $= 1 \times 0.14 = 0.140 \ M$.
$(d) \ MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$, so $i = 3$. Product $= 3 \times 0.04 = 0.120 \ M$.
Comparing the products $(i \times m)$: $(b) \ 0.180 > (a) \ 0.150 > (c) \ 0.140 > (d) \ 0.120$.
Since freezing point decreases as the product $i \times m$ increases, the order of decreasing freezing points is $(c) > (d) > (a) > (b)$.

(B) 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.
$\Delta T_f = T_f^\circ - T_f = 0 \ ^oC - (-4.02 \ ^oC) = 4.02 \ K$.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{x}{1.2 \ kg}$.
Substituting the values into the formula: $4.02 = 1.86 \times \frac{x}{1.2}$.
Solving for $x$: $x = \frac{4.02 \times 1.2}{1.86} = \frac{4.824}{1.86} \approx 2.59 \ \text{moles}$.

(C) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ and molality $m = 0.05 \ molal$.
$\Delta T_f = 1.86 \times 0.05 = 0.093 \ K$
Since the freezing point of pure water is $0 \ ^oC$,the freezing point of the solution is:
$T_f = T_f^\circ - \Delta T_f = 0 \ ^oC - 0.093 \ ^oC = - 0.093 \ ^oC$

(D) The freezing point of a solution is determined by the colligative property known as depression of freezing point. $\Delta T_f = i \times K_f \times m$. Since calcium nitrate $(Ca(NO_3)_2)$ is a solute,its addition to water lowers the freezing point of the solvent. Therefore,the freezing point of the solution will be less than the freezing point of pure water $(0\,^{\circ}C)$.

(C) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62 \ g \ mol^{-1}$.
Using the formula for depression in freezing point: $\Delta T_f = K_f \times m$,where $m$ is the molality.
$\Delta T_f = 9.3 \ K$.
$9.3 = 1.86 \times \frac{50 / 62}{W / 1000}$,where $W$ is the mass of water remaining in liquid state.
$9.3 = \frac{1.86 \times 50 \times 1000}{62 \times W}$.
$W = \frac{1.86 \times 50 \times 1000}{62 \times 9.3} = 161.29 \ g$.
The amount of ice separated is the initial mass of water minus the mass of water remaining in liquid state.
$\text{Ice separated} = 200 \ g - 161.29 \ g = 38.71 \ g$.

(A) $T_1 =$ Freezing point of pure solvent
$T_2 =$ Freezing point of solution
$\Delta T_f = T_1 - T_2 = i \times K_f \times m$
For urea,van't Hoff factor $i = 1$.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{6 \ g / 60 \ g \ mol^{-1}}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.
$\Delta T_f = 1 \times 2 \ kg \ K \ mol^{-1} \times 0.1 \ mol \ kg^{-1} = 0.2 \ K$.
Since the difference in temperature is the same in $^\circ C$ and $K$,$T_1 - T_2 = 0.2 \ ^\circ C$.

(C) The freezing point of a solution is lowered by the addition of a solute (depression in freezing point).
$CaCl_2$ is preferred because the eutectic mixture of $CaCl_2$ and $H_2O$ has a much lower freezing point $(-55 \ ^oC)$ compared to the eutectic mixture of $NaCl$ and $H_2O$ $(-18 \ ^oC)$.
This allows $CaCl_2$ to effectively melt ice at significantly lower temperatures.

(D) For the same freezing point,the molality $(m)$ of both aqueous solutions must be equal,assuming both are non-electrolytes.
$m_A = m_B$
The formula for molality is $m = \frac{\text{mass of solute (g)} \times 1000}{\text{molecular weight of solute} \times \text{mass of solvent (g)}}$.
For a $4 \%$ solution of $A$,mass of $A = 4 \ g$ and mass of water $= 96 \ g$.
For a $10 \%$ solution of $B$,mass of $B = 10 \ g$ and mass of water $= 90 \ g$.
Substituting the values: $\frac{4 \times 1000}{60 \times 96} = \frac{10 \times 1000}{M_B \times 90}$.
Simplifying the equation: $\frac{4}{60 \times 96} = \frac{10}{M_B \times 90}$.
$M_B = \frac{10 \times 60 \times 96}{4 \times 90} = \frac{57600}{360} = 160$.

(B) At the freezing point of a solution,the liquid phase of the solvent is in equilibrium with the solid phase of the pure solvent.
This is because,upon cooling,only the solvent molecules crystallize out as solid,while the solute remains in the remaining liquid solution.
Therefore,the equilibrium is represented as: Solvent$_{(l)}$ $\rightleftharpoons$ Solvent$_{(s)}$.

(B) 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: $\Delta T_f = 6 \ K$,$K_f = 1.85 \ K \ kg \ mol^{-1}$,$W_A = 4 \ kg$,$M_B = 62 \ g \ mol^{-1}$.
Substituting the values in the formula: $\Delta T_f = K_f \times \frac{w_B}{M_B \times W_A}$
$6 = \frac{1.85 \times w_B}{62 \times 4}$
$w_B = \frac{6 \times 62 \times 4}{1.85}$
$w_B = \frac{1488}{1.85} \approx 804.32 \ g$
Thus,the amount of ethylene glycol required is approximately $804.3 \ g$.

(C) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ and molality $m = 0.05 \ molal$.
$\Delta T_f = 1.86 \times 0.05 = 0.093 \ K$.
Since the freezing point of pure water is $0 \ ^oC$,the freezing point of the solution is $T_f = T_f^0 - \Delta T_f$.
$T_f = 0 - 0.093 = -0.093 \ ^oC$.

(B) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62 \ g \ mol^{-1}$.
Given: $\Delta T_f = 9.3 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,mass of solute $(w)$ = $40 \ g$.
Using the formula $\Delta T_f = \frac{1000 \times K_f \times w}{M \times W}$,where $W$ is the mass of the solvent remaining in liquid state.
$9.3 = \frac{1000 \times 1.86 \times 40}{62 \times W}$
$9.3 = \frac{74400}{62 \times W}$
$W = \frac{74400}{62 \times 9.3} = \frac{74400}{576.6} \approx 129.03 \ g$.
The amount of ice separated = (Initial mass of water) - (Mass of water remaining in liquid state).
Ice separated = $400 - 129.03 = 270.97 \ g$.

(D) When a solution containing a non-volatile solute freezes,the solid phase that separates out is pure solid solvent.
Therefore,the equilibrium is established between the pure solid solvent and the solvent present in the liquid solution.
The equilibrium can be represented as: $\text{solid solvent} \rightleftharpoons \text{solvent in solution}$.

(A) The molar mass of ethylene glycol $(C_2H_6O_2)$ is $62\, g\, mol^{-1}$.
Given $28\%$ by mass solution means $28\, g$ of ethylene glycol in $72\, g$ of water.
Molality $(m)$ $= \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{28}{62 \times 0.072} \approx 6.27\, mol\, kg^{-1}$.
Depression in freezing point $\Delta T_f = K_f \times m = 1.86 \times 6.27 \approx 11.66\, K$.
The freezing point of the solution $= 0 - 11.66 = -11.66\, ^{\circ}C$.
Since the freezing point of the solution $(-11.66\, ^{\circ}C)$ is lower than the ambient temperature $(-11\, ^{\circ}C)$,the solution will not freeze.
Therefore,it is suitable for a car radiator.

(D) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$,where $i$ is the van't Hoff factor,$K_f$ is the cryoscopic constant,and $m$ is the molality.
Since $K_f$ and $m$ are constant for all solutions,the depression in freezing point depends on the van't Hoff factor $(i)$.
Freezing point $T_f = T_f^0 - \Delta T_f$. To have the maximum freezing point,the depression $\Delta T_f$ must be minimum,which means $i$ must be minimum.
Let us calculate $i$ for each solute:
$(A)$ $CH_3COONa \rightarrow CH_3COO^- + Na^+$,$i = 2$
$(B)$ $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,$i = 3$
$(C)$ $Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}$,$i = 4$
$(D)$ $C_{12}H_{22}O_{11}$ (sucrose) is a non-electrolyte,$i = 1$
Since $C_{12}H_{22}O_{11}$ has the smallest $i$ value,it will have the minimum depression in freezing point and thus the maximum freezing point.

(B) The depression in freezing point is given by $\Delta T_f = K_f \times \sum m_i$,where $m_i$ is the molality of each solute.
For $5\%$ urea (molar mass $= 60 \ g \ mol^{-1}$) in $95 \ g$ water: $m_{urea} = \frac{5 \times 1000}{60 \times 95} \approx 0.877 \ mol \ kg^{-1}$.
For $10\%$ glucose (molar mass $= 180 \ g \ mol^{-1}$) in $90 \ g$ water: $m_{glucose} = \frac{10 \times 1000}{180 \times 90} \approx 0.617 \ mol \ kg^{-1}$.
Total $\Delta T_f = 1.86 \times (0.877 + 0.617) = 1.86 \times 1.494 \approx 2.78 \ ^oC$.
Since $T_f = T_f^{\circ} - \Delta T_f$ and $T_f^{\circ} = 0 \ ^oC$,the freezing point is $0 - 2.78 = -2.78 \ ^oC$.

(A) At the freezing point of a solution,the liquid solvent and the solid solvent are in equilibrium.
During the freezing process,only the solvent molecules transition from the liquid phase to the solid phase.
The vapour pressure of the solid solvent and the liquid solvent must be equal at the freezing point for the system to be in equilibrium.
Therefore,the correct equilibrium is between the liquid solvent and the solid solvent.

(B) The depression in freezing point $(\Delta T_f)$ is calculated as:
$\Delta T_f = T_f^{\circ} - T_f = 278.82 \ K - 276.82 \ K = 2.00 \ K$.
The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given $m = 0.25 \ mol \ kg^{-1}$ and $\Delta T_f = 2.00 \ K$.
Substituting the values: $2.00 \ K = K_f \times 0.25 \ mol \ kg^{-1}$.
Therefore,$K_f = \frac{2.00}{0.25} = 8 \ K \ kg \ mol^{-1}$.
Thus,the correct option is $B$.

(C) The depression of freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \times m$,where $K_f$ is the cryoscopic constant (molal depression constant) and $m$ is the molality of the solution.
Therefore,the cryoscopic constant $K_f$ is defined as the ratio of the depression of freezing point $(\Delta T_f)$ to the molality $(m)$ of the solution: $K_f = \frac{\Delta T_f}{m}$.

(A) The freezing point of a substance is defined as the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase. In the case of a solution,when it reaches its freezing point,the liquid solvent is in equilibrium with the solid solvent. The solute remains in the liquid phase.

(A) The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 144 + 22 + 176 = 342\,g/mol$.
Given mass of solute $(w_2) = 342\,g$.
Given mass of solvent $(w_1) = 1000\,g = 1\,kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{342/342}{1} = 1\,m$.
The depression in freezing point $(\Delta T_f)$ is given by $\Delta T_f = K_f \times m = 1.86 \times 1 = 1.86\,^{\circ}C$.
Freezing point of solution $(T_f) = T_f^{\circ} - \Delta T_f = 0 - 1.86 = -1.86\,^{\circ}C$.

(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2$ is the mass of solute (urea),$M_2$ is the molar mass of urea $(60 \ g/mol)$,and $w_1$ is the mass of solvent (water).
Given: $\Delta T_f = 0.186 \ ^oC$,$K_f = 1.86 \ ^oC/m$,$w_1 = 500 \ g$ (since density of water is $1 \ g/mL$,$500 \ mL = 500 \ g$).
Substituting the values: $0.186 = 1.86 \times \frac{w_2 \times 1000}{60 \times 500}$.
$0.186 = 1.86 \times \frac{w_2 \times 2}{60}$.
$0.186 = 1.86 \times \frac{w_2}{30}$.
$w_2 = \frac{0.186 \times 30}{1.86} = 0.1 \times 30 = 3 \ g$.

(B) When a solution is cooled,the solvent (water) begins to freeze first because the freezing point of the solution is lower than that of the pure solvent. As the solvent freezes,the concentration of the solute (glucose) in the remaining liquid increases,further lowering the freezing point. Therefore,pure $H_2O$ crystals separate out first.

(D) The depression in freezing point 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. The freezing point is lowest when the depression in freezing point $(\Delta T_f)$ is maximum. This depends on the value of $i \times m$.
For $A$: $KNO_3 \rightarrow K^+ + NO_3^-$,so $i = 2$. $i \times m = 2 \times 0.08 = 0.16$.
For $B$: $FeCl_3 \rightarrow Fe^{3+} + 3Cl^-$,so $i = 4$. $i \times m = 4 \times 0.03 = 0.12$.
For $C$: $Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$,so $i = 3$. $i \times m = 3 \times 0.05 = 0.15$.
For $D$: $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$. $i \times m = 5 \times 0.04 = 0.20$.
Since the value of $i \times m$ is highest for $D$ $(0.20)$,it will have the maximum depression in freezing point,and thus the lowest freezing point.

(C) Given: $K_f = 7.00$,$W_2 = 0.072 \ g-atom \times 32 \ g/mol = 2.304 \ g$,$W_1 = 100 \ g$,$\Delta T_f = 0.84 \ ^\circ C$.
Using the formula: $\Delta T_f = K_f \times m = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1}$.
$0.84 = 7.00 \times \frac{2.304 \times 1000}{M_2 \times 100}$.
$M_2 = \frac{7.00 \times 2.304 \times 10}{0.84} = \frac{161.28}{0.84} = 192 \ g/mol$.
Since the atomic mass of sulfur is $32 \ g/mol$,the number of atoms in the molecule is $n = \frac{192}{32} = 6$.
Therefore,the molecular formula is $S_6$.

(C) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality.
For a $5\%$ solution (by mass),the molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$.
Since the percentage by mass is the same $(5\%)$,the molality is inversely proportional to the molar mass $(M_2)$ of the solute.
$\Delta T_f \propto \frac{1}{M_2}$.
For cane sugar $(M_1 = 342 \ g/mol)$: $\Delta T_{f1} = 273.15 - 271 = 2.15 \ K$.
For glucose $(M_2 = 180 \ g/mol)$: $\Delta T_{f2} = \Delta T_{f1} \times \frac{M_1}{M_2}$.
$\Delta T_{f2} = 2.15 \times \frac{342}{180} = 2.15 \times 1.9 = 4.085 \ K$.
The freezing point of the glucose solution $= 273.15 - 4.085 = 269.065 \ K \approx 269.07 \ K$.

(A) The depression in freezing point $\Delta T_f$ is directly proportional to the van't Hoff factor $i$ for solutions of the same molality: $\Delta T_f = i \times K_f \times m$.
Since the molality $m$ is the same for all,the solution with the highest value of $i$ will have the greatest depression in freezing point and thus the lowest freezing point.
$1$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ $(i = 5)$
$2$. $C_6H_{12}O_6$ (glucose) is a non-electrolyte $(i = 1)$
$3$. $KCl \rightarrow K^{+} + Cl^{-}$ $(i = 2)$
$4$. $C_{12}H_{22}O_{11}$ (sucrose) is a non-electrolyte $(i = 1)$
Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i = 5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.
Here,$\Delta T_{f} = 0.2 \ K$,$K_{f} = 2 \ K \ kg \ mol^{-1}$,and for $NaCl$,the van't Hoff factor $i = 2$.
The molality $m = \frac{w / 58.5}{0.6 \ kg}$,where $w$ is the mass of $NaCl$ in grams.
Substituting the values: $0.2 = 2 \times 2 \times \frac{w}{58.5 \times 0.6}$.
$0.2 = 4 \times \frac{w}{35.1}$.
$w = \frac{0.2 \times 35.1}{4} = 1.755 \ g$.

(N/A) The depression in freezing point is calculated using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
$1.$ Calculate the moles of ethylene glycol: $\text{Moles} = \frac{45 \ g}{62 \ g \ mol^{-1}} = 0.726 \ mol$.
$2.$ Calculate the mass of water in $kg$: $\text{Mass} = \frac{600 \ g}{1000 \ g \ kg^{-1}} = 0.6 \ kg$.
$3.$ Calculate the molality $(m)$: $m = \frac{0.726 \ mol}{0.6 \ kg} = 1.21 \ mol \ kg^{-1}$.
$4.$ Calculate the freezing point depression $(\Delta T_{f})$: $\Delta T_{f} = 1.86 \ K \ kg \ mol^{-1} \times 1.21 \ mol \ kg^{-1} = 2.25 \ K$.
$5.$ Calculate the freezing point of the solution: $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 273.15 \ K - 2.25 \ K = 270.90 \ K$.