Depression of freezing point of the solvent Questions in English

(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality.
$m = \frac{W_2 \times 1000}{M_2 \times W_1} = \frac{1 \ g \times 1000}{60 \ g \ mol^{-1} \times 100 \ g} = \frac{1}{6} \ mol \ kg^{-1}$.
Given $\Delta T_{f} = 0.3 \ K$.
Using $\Delta T_{f} = K_{f} \times m$:
$0.3 \ K = K_{f} \times \frac{1}{6} \ mol \ kg^{-1}$.
$K_{f} = 0.3 \times 6 = 1.8 \ K \ kg \ mol^{-1}$.

(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{W_2 \times 1000}{M_2 \times W_1}$,where $W_2$ is the mass of the solute,$M_2$ is the molar mass of the solute,and $W_1$ is the mass of the solvent in grams.
Substituting the given values: $W_2 = 3.2 \ g$,$M_2 = 128 \ g \ mol^{-1}$,$W_1 = 80 \ g$,and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.
$\Delta T_{f} = \frac{4.8 \ K \ kg \ mol^{-1} \times 3.2 \ g \times 1000 \ g \ kg^{-1}}{128 \ g \ mol^{-1} \times 80 \ g}$.
$\Delta T_{f} = \frac{15360}{10240} \ K = 1.5 \ K$.

(B) Given: Mass of solute $(W_2) = 4 \,g$, Molar mass of solute $(M_2) = 126 \,g \,mol^{-1}$, Volume of water $= 80 \,mL$.
Since the density of water is $1 \,g/mL$, the mass of solvent $(W_1) = 80 \,g$.
The formula for depression in freezing point is $\Delta T_f = \frac{1000 \times K_f \times W_2}{M_2 \times W_1}$.
Substituting the values: $\Delta T_f = \frac{1000 \times 1.86 \times 4}{126 \times 80}$.
$\Delta T_f = \frac{7440}{10080} \approx 0.738 \,K$, which rounds to $0.74 \,K$.

(C) Given: Mass of solute $(W_2)$ = $15 \ g$,Volume of water = $200 \ mL$,so Mass of solvent $(W_1)$ = $200 \ g$ (assuming density = $1 \ g/mL$).
Depression in freezing point $(\Delta T_f)$ = $0.75 \ K$.
Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$.
The formula for molar mass $(M_2)$ is:
$M_2 = \frac{1000 \times K_f \times W_2}{\Delta T_f \times W_1}$
Substituting the values:
$M_2 = \frac{1000 \times 1.86 \times 15}{0.75 \times 200}$
$M_2 = \frac{27900}{150} = 186 \ g \ mol^{-1}$.

(A) The cryoscopic constant $(K_f)$,also known as the molal freezing point depression constant,is defined by the equation: $\Delta T_f = K_f \times m$,where $\Delta T_f$ is the depression in freezing point $(K)$ and $m$ is the molality $(mol \ kg^{-1})$.
Rearranging for $K_f$,we get: $K_f = \frac{\Delta T_f}{m} = \frac{K}{mol \ kg^{-1}} = K \ kg \ mol^{-1}$.

(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$.
Here,$\Delta T_f = 0.2 \ K$,$K_f = 1.71 \ K \ kg \ mol^{-1}$,$M_B = 60 \ g \ mol^{-1}$,and $W_A = 98 \ g$.
The molality $m$ is given by $\frac{W_B \times 1000}{M_B \times W_A}$.
Substituting the values: $0.2 = 1.71 \times \frac{W_B \times 1000}{60 \times 98}$.
Solving for $W_B$: $W_B = \frac{0.2 \times 60 \times 98}{1.71 \times 1000} = \frac{1176}{1710} \approx 0.687 \ g$.

(A) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2 = 5 \ g$ (solute mass),$w_1 = 70 \ g$ (solvent mass),and $M_2$ is the molar mass of the solute.
Substituting the values: $2.5 = 3.5 \times \frac{5 \times 1000}{M_2 \times 70}$.
$M_2 = \frac{3.5 \times 5 \times 1000}{2.5 \times 70} = \frac{17500}{175} = 100 \ g \ mol^{-1}$.

(C) The formula for freezing point depression is given by: $\Delta T_{f} = K_{f} \times m$
Where $\Delta T_{f}$ is the freezing point depression,$K_{f}$ is the freezing point depression constant,and $m$ is the molality.
Given: $\Delta T_{f} = 3.6 \ K$ and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.
Substituting the values in the formula: $3.6 = 4.8 \times m$
$m = \frac{3.6}{4.8} = 0.75 \ mol \ kg^{-1}$.

(A) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
The mass of the solvent $(W_A)$ is calculated as: $W_A = \text{density} \times \text{volume} = 0.8 \ g \ mL^{-1} \times 100 \ mL = 80 \ g$.
Substituting the given values into the formula: $\Delta T_f = K_f \times \frac{W_B}{M_B} \times \frac{1000}{W_A(g)}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times \frac{1000}{80}$.
$0.75 = 5 \times \frac{1.5}{M_B} \times 12.5$.
$M_B = \frac{5 \times 1.5 \times 12.5}{0.75} = \frac{93.75}{0.75} = 125 \ g \ mol^{-1}$.

(D) The depression in freezing point $(\Delta T_f)$ is given by the formula: $\Delta T_f = K_f \cdot m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 \text{ in grams})$: $m = \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Substituting this into the depression in freezing point equation: $\Delta T_f = K_f \cdot \frac{W_2 \cdot 1000}{M_2 \cdot W_1}$.
Rearranging the formula to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$.

(D) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$.
Given:
Molality $(m) = 0.25 \ mol \ kg^{-1}$
Cryoscopic constant $(K_{f}) = 4.0 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_{f} = 4.0 \times 0.25 = 1.0 \ K$
Thus,the freezing point depression is $1.0 \ K$.

(B) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water and molality $m = 0.05 \ m$.
$\Delta T_f = 1.86 \times 0.05 = 0.093 \ K$.
The freezing point of the solution $T_f$ is calculated as: $T_f = T_f^{\circ} - \Delta T_f$.
Since the freezing point of pure water $T_f^{\circ} = 273 \ K$,we have: $T_f = 273 - 0.093 = 272.907 \ K \approx 272.9 \ K$.

(C) The depression in freezing point $(\Delta T_f)$ is directly proportional to the molality $(m)$ of the solution.
$\Delta T_f = K_f \times m$
Where $K_f$ is the cryoscopic constant (molal depression constant).
Rearranging the equation to solve for $K_f$ gives:
$K_f = \frac{\Delta T_f}{m}$
Therefore,the correct option is $C$.

(D) The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.8 \ g / 64 \ g \ mol^{-1}}{43 \ g / 1000} = \frac{0.0125 \ mol}{0.043 \ kg} \approx 0.2907 \ mol \ kg^{-1}$.
Given $\Delta T_f = 0.34 \ K$,we have $0.34 = K_f \times 0.2907$.
$K_f = \frac{0.34}{0.2907} \approx 1.169 \ K \ kg \ mol^{-1}$.
Rounding to two decimal places,$K_f \approx 1.17 \ K \ kg \ mol^{-1}$.

(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Given that $\Delta T_{f} = K_{f}$,we can substitute this into the equation: $K_{f} = K_{f} \times m$.
Dividing both sides by $K_{f}$,we get $m = 1$.
Therefore,the concentration of the solution must be $1 \ m$ (molal).

(A) The formula for freezing point depression is $\Delta T_f = K_f \cdot m$.
Given: $\Delta T_f = 3 \ K$ and $K_f = 5 \ K \ kg \ mol^{-1}$.
Substituting the values into the equation: $3 = 5 \cdot m$.
Solving for molality $(m)$: $m = \frac{3}{5} = 0.6 \ m$.

(A) The formula for depression in freezing point is $\Delta T_f = K_f \cdot m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{5 \ g / 180 \ g \ mol^{-1}}{0.1 \ kg} = \frac{5}{18} \ mol \ kg^{-1} \approx 0.2778 \ mol \ kg^{-1}$.
Given $\Delta T_f = 2.15 \ K$.
Substituting the values into the formula: $2.15 = K_f \cdot (5/18)$.
$K_f = \frac{2.15 \times 18}{5} = 7.74 \ K \ kg \ mol^{-1}$.

(A) The formula for freezing point depression is $\Delta T_{f} = K_{f} \cdot m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{6 / M}{100 / 1000} = \frac{6}{M} \times 10 = \frac{60}{M} \ mol \ kg^{-1}$.
Given $\Delta T_{f} = 0.93 \ K$ and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.93 = 1.86 \times \frac{60}{M}$.
$M = \frac{1.86 \times 60}{0.93} = 2 \times 60 = 120 \ g \ mol^{-1}$.

(C) The molal depression constant $(K_f)$ is defined as the depression in freezing point for a $1 \ molal$ solution.
Since $K_f = \frac{\Delta T_f}{m}$,the unit involves a temperature difference.
$A$ temperature difference of $1^{\circ} C$ is equivalent to a temperature difference of $1 \ K$.
Therefore,the numerical value of the molal depression constant remains the same when expressed in $K \ kg \ mol^{-1}$.
Thus,$2.77^{\circ} C \ kg \ mol^{-1} = 2.77 \ K \ kg \ mol^{-1}$.

(B) The depression in freezing point $( \Delta T_{f} )$ is defined as the difference between the freezing point of the pure solvent $( T_{f}^{\circ} )$ and the freezing point of the solution $( T_{f} )$.
Therefore,the correct formula is $ \Delta T_{f} = T_{f}^{\circ} - T_{f} $.

(A) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \cdot m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{5 / 342}{0.1 \ kg} = \frac{50}{342} \ mol \ kg^{-1}$.
Now,substitute the values into the depression in freezing point equation:
$2.15 = K_{f} \cdot \frac{50}{342}$.
Solving for $K_{f}$:
$K_{f} = \frac{2.15 \times 342}{50} = 14.7 \ K \ kg \ mol^{-1}$.

(D) The depression in freezing point is given by the formula: $\Delta T_{f} = K_{f} \cdot m$
Substituting the given values: $\Delta T_{f} = 1.86 \ K \ kg \ mol^{-1} \times 1 \ mol \ kg^{-1} = 1.86 \ K$
Since the change in temperature in Celsius is the same as in Kelvin,$\Delta T_{f} = 1.86^{\circ} C$
The freezing point of the solution is calculated as: $T_{f} = T_{f}^{\circ} - \Delta T_{f}$
$T_{f} = 0^{\circ} C - 1.86^{\circ} C = -1.86^{\circ} C$

(A) Since both solutions freeze at the same temperature,their freezing point depression $(\Delta T_f)$ is equal.
For dilute aqueous solutions,$\Delta T_f = K_f \times m$,where $m$ is the molality.
Since $K_f$ is the same for both,$m_1 = m_2$.
Given $1 \ dm^3$ of water $\approx 1 \ kg$ of water.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For urea: $m_1 = \frac{6 \ g / 60 \ g \cdot mol^{-1}}{1 \ kg} = 0.1 \ mol \cdot kg^{-1}$.
For solute $A$: $m_2 = \frac{9 \ g / M_A}{1 \ kg} = \frac{9}{M_A} \ mol \cdot kg^{-1}$.
Equating $m_1 = m_2$: $0.1 = \frac{9}{M_A}$.
Therefore,$M_A = \frac{9}{0.1} = 90 \ g \cdot mol^{-1}$.

(A) Mass of water $= 0.5 \ dm^{3} = 0.5 \ kg$
Moles of urea $= \frac{30 \ g}{60 \ g \ mol^{-1}} = 0.5 \ mol$
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.5 \ mol}{0.5 \ kg} = 1 \ mol \ kg^{-1}$
The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$
Given $\Delta T_{f} = 0.15 \ K$
$K_{f} = \frac{\Delta T_{f}}{m} = \frac{0.15 \ K}{1 \ mol \ kg^{-1}} = 0.15 \ K \ kg \ mol^{-1}$

(D) The depression in freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
$\Delta T_{f} = T^{\circ} - T$
Rearranging the equation to solve for the freezing point of the pure solvent $(T^{\circ})$:
$T^{\circ} = \Delta T_{f} + T$
Where:
$T^{\circ} = \text{freezing point of pure solvent}$
$T = \text{freezing point of solution}$
$\Delta T_{f} = \text{depression in freezing point}$

(D) The formula for depression in freezing point is $\Delta T_{f} = k_{f} \times m$.
First,calculate the molality $(m)$:
$m = \frac{\text{mass of solute} \times 1000}{\text{molar mass of solute} \times \text{mass of solvent in g}} = \frac{1.5 \times 1000}{60 \times 250} = 0.1 \ m$.
Given $\Delta T_{f} = 0.01 \ ^{\circ}C$.
Substituting the values into the formula:
$0.01 = k_{f} \times 0.1$.
Therefore,$k_{f} = \frac{0.01}{0.1} = 0.1 \ ^{\circ}C \ kg \ mol^{-1}$.

(A) $1$. Molar Mass Calculation:
The molecular formula of ethylene glycol is $C_2H_6O_2$.
Molar Mass $(M_2) = (2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g \ mol^{-1}$.
$2$. Setup:
Let the mass of ethylene glycol (solute) be $w_2 \ g$.
Total mass of solution = $645 \ g$.
Mass of water (solvent),$w_1 = (645 - w_2) \ g$.
Given: $\Delta T_f = 2.25 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$.
$3$. Calculation:
Using the freezing point depression formula: $\Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$.
Substituting the values: $2.25 = \frac{1.86 \times w_2 \times 1000}{62 \times (645 - w_2)}$.
$2.25 = \frac{30 \times w_2}{645 - w_2}$.
$2.25(645 - w_2) = 30w_2$.
$1451.25 - 2.25w_2 = 30w_2$.
$1451.25 = 32.25w_2$.
$w_2 = \frac{1451.25}{32.25} = 45 \ g$.

(B) The freezing point depression $\Delta T_f$ is a colligative property,which depends on the molality $(m)$ of the solution. $\Delta T_f = K_f \times m$. For the freezing points to be equal,the molality of the two solutions must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For option $B$:
Moles of urea $= \frac{6 \ g}{60 \ g/mol} = 0.1 \ mol$.
Moles of glucose $= \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
Since the mass of the solvent $(100 \ g \ H_2O)$ is the same for both,and the number of moles of solute is the same $(0.1 \ mol)$,the molality of both solutions is identical.
Therefore,they will have the same freezing point.

(D) Given: $w_2 = 1.00 \text{ g}$,$w_1 = 50 \text{ g}$,$\Delta T_f = 0.40 \text{ K}$,$K_f = 5.12 \text{ K kg mol}^{-1}$.
Using the formula: $\Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$
Rearranging for $M_2$: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Substituting the values: $M_2 = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50}$
$M_2 = \frac{5120}{20} = 256 \text{ g mol}^{-1}$.

(B) The freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$.
The freezing point of the solution is $T_f = T_f^\circ - \Delta T_f$.
For the highest freezing point,$\Delta T_f$ must be minimum,which implies that the product $i \cdot M$ must be minimum.
$(A)$ For $0.1 \ M$ Sucrose,$i = 1$,$i \cdot M = 1 \times 0.1 = 0.1$.
$(B)$ For $0.01 \ M \ NaCl$,$i = 2$,$i \cdot M = 2 \times 0.01 = 0.02$.
$(C)$ For $0.1 \ M \ NaCl$,$i = 2$,$i \cdot M = 2 \times 0.1 = 0.2$.
$(D)$ For $0.01 \ M \ Na_2SO_4$,$i = 3$,$i \cdot M = 3 \times 0.01 = 0.03$.
Since $0.01 \ M \ NaCl$ has the lowest $i \cdot M$ value,it has the highest freezing point.

(A) In countries nearer to the polar region,the roads are sprinkled with $CaCl_{2}$ because $CaCl_{2}$ acts as a freezing point depressant.
It lowers the freezing point of water,which helps in melting the ice on the roads,thereby preventing ice formation and minimizing the wear and tear of the roads.

(C) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Since both solutions have the same freezing point and the same amount of solvent ($1 \ kg$ of water),their molalities must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
For urea: $\text{moles} = \frac{7.5 \ g}{60 \ g \ mol^{-1}} = 0.125 \ mol$.
For solute $X$: $\text{moles} = \frac{15 \ g}{M_X}$,where $M_X$ is the molar mass of $X$.
Equating the moles (since solvent mass is the same): $0.125 = \frac{15}{M_X}$.
$M_X = \frac{15}{0.125} = 120 \ g \ mol^{-1}$.

(B) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w_{solute} \times 1000}{M_{solute} \times w_{solvent(g)}}$.
Given: $\Delta T_{f} = 0.64 \ K$,$w_{solute} = 1.95 \ g$,$w_{solvent} = 100 \ g$,$K_{f} = 5.12 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.64 = 5.12 \times \frac{1.95 \times 1000}{M_{solute} \times 100}$.
$0.64 = 5.12 \times \frac{19.5}{M_{solute}}$.
$M_{solute} = \frac{5.12 \times 19.5}{0.64}$.
$M_{solute} = 8 \times 19.5 = 156 \ g \ mol^{-1}$.

(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273.16 \ K - 272.814 \ K = 0.346 \ K$.
The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality.
Molality $m = \frac{w_A \times 1000}{M_A \times w_{solvent}}$,where $w_A = 0.2 \ g$ and $w_{solvent} = 21.5 \ g$.
Substituting the values: $0.346 = 1.86 \times \frac{0.2 \times 1000}{M_A \times 21.5}$.
$M_A = \frac{1.86 \times 0.2 \times 1000}{0.346 \times 21.5} = \frac{372}{7.439} \approx 50 \ g \ mol^{-1}$.
Thus,the molar mass of solute '$A$' is $50 \ g \ mol^{-1}$.

(D) The formula for depression in freezing point is $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality.
Molality $m = \frac{W_{solute} \times 1000}{M_{solute} \times W_{solvent} \text{ (in g)}}$.
For $XY$: $5.333 = 2 \times \frac{10 \times 1000}{M_{XY} \times 50} \implies M_{XY} = \frac{20000}{5.333 \times 50} \approx 75 \ g/mol$.
So,$X + Y = 75$ (Equation $1$).
For $XY_3$: $2.2857 = 2 \times \frac{10 \times 1000}{M_{XY_3} \times 50} \implies M_{XY_3} = \frac{20000}{2.2857 \times 50} \approx 175 \ g/mol$.
So,$X + 3Y = 175$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(X + 3Y) - (X + Y) = 175 - 75 \implies 2Y = 100 \implies Y = 50 \ u$.
Substituting $Y = 50$ in Equation $1$: $X + 50 = 75 \implies X = 25 \ u$.
The atomic weights are $X = 25 \ u$ and $Y = 50 \ u$.

(A) The molal depression constant $(K_{f})$,also known as the cryoscopic constant,is defined by the relation $K_{f} = \frac{R \cdot M_{solvent} \cdot T_{f}^{2}}{1000 \cdot \Delta H_{fus}}$.
Since $R$,$M_{solvent}$ (molar mass of solvent),and $\Delta H_{fus}$ (enthalpy of fusion of solvent) are properties specific to the solvent,the value of $K_{f}$ depends solely on the nature of the solvent.

(D) The depression in freezing point is given by $\Delta T_f = m \times K_f$,where $m$ is the molality of the solution.
Molality $m = \frac{n_2}{w_1 (\text{in } kg)}$,where $n_2$ is the moles of solute and $w_1$ is the mass of solvent in $kg$.
Given the mole fraction of solute $x_2 = 0.01$,we have $x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = 0.01$,where $n_1$ is the moles of water.
For $1 \ kg$ of water $(w_1 = 1 \ kg)$,$n_1 = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.
Thus,$n_2 = 0.01 \times 55.55 \ mol = 0.5555 \ mol$.
Now,$m = \frac{0.5555 \ mol}{1 \ kg} = 0.5555 \ mol \ kg^{-1}$.
$\Delta T_f = 0.5555 \ mol \ kg^{-1} \times 1.86 \ K \ kg \ mol^{-1} = 1.0332 \ K \approx 1.043 \ K$ (considering standard approximations).

(B) Depression in freezing point is a colligative property,which depends on the molality of the solution.
$\Delta T_{f} = K_{f} \times m$
Where $\Delta T_{f} = 1.0 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = \frac{x / 78}{0.5 \ kg}$.
Substituting the values: $1.0 = 1.86 \times \frac{x}{78 \times 0.5}$.
$1.0 = 1.86 \times \frac{x}{39}$.
$x = \frac{39}{1.86} \approx 20.96 \ g$.

(D) Given: $n = 0.05 \ mol$
Weight of solvent $(W) = 500 \ g = 0.5 \ kg$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
Depression in freezing point $(\Delta T_f) = ?$
Molality $(m) = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in } kg} = \frac{0.05}{0.5} = 0.1 \ m$
Using the formula: $\Delta T_f = m \times K_f$
$\Delta T_f = 0.1 \times 1.86 = 0.186 \ K$

(C) The freezing point depression is a colligative property,given by the formula $\Delta T_f = i \cdot K_f \cdot m$,where $i$ is the van't Hoff factor and $m$ is the molality.
Freezing point $T_f = T_f^0 - \Delta T_f$. To have the highest freezing point,the depression $\Delta T_f$ must be the minimum.
$1$. For $0.1 \ mol \ KCl$: $i = 2$,$m = 0.1$,so $\Delta T_f \propto 2 \times 0.1 = 0.2$.
$2$. For $0.1 \ mol \ K_2SO_4$: $i = 3$,$m = 0.1$,so $\Delta T_f \propto 3 \times 0.1 = 0.3$.
$3$. For $0.1 \ mol$ Urea: $i = 1$,$m = 0.1$,so $\Delta T_f \propto 1 \times 0.1 = 0.1$.
$4$. For $30 \ g$ Glucose $(M = 180 \ g/mol)$: $m = \frac{30}{180} = 0.167 \ mol/kg$,$i = 1$,so $\Delta T_f \propto 1 \times 0.167 = 0.167$.
Comparing the values,the minimum $\Delta T_f$ is for $0.1 \ mol$ Urea. Thus,it has the highest freezing point.

(C) Given: $\text{Mass of urea} = 0.6 \ g$
$\text{Molar mass of urea} = 60 \ g \ mol^{-1}$
$K_f \text{ of benzene} = 4.0 \ K \ kg \ mol^{-1}$
$\text{Volume of benzene} = 100 \ mL$.
Assuming the density of benzene is $1 \ g \ mL^{-1}$,the mass of the solvent is $100 \ g = 0.1 \ kg$.
$\text{Moles of urea} = \frac{0.6 \ g}{60 \ g \ mol^{-1}} = 0.01 \ mol$.
$\text{Molality } (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.01 \ mol}{0.1 \ kg} = 0.1 \ mol \ kg^{-1}$.
$\text{Freezing point depression } (\Delta T_f) = K_f \times m = 4.0 \ K \ kg \ mol^{-1} \times 0.1 \ mol \ kg^{-1} = 0.4 \ K$.
Therefore,option $C$ is correct.

(C) Depression in freezing point is given as $\Delta T_f = i \times K_f \times m$.
Here,$i = 1$ (for glucose,a non-electrolyte).
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Molality $(m)$ is calculated as:
$m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1}{\text{mass of solvent in kg}} = \frac{1.8 \ g}{180 \ g \ mol^{-1}} \times \frac{1}{0.1 \ kg} = 0.01 \ mol \times 10 \ kg^{-1} = 0.1 \ m$.
Now,$\Delta T_f = 1 \times 1.86 \ K \ kg \ mol^{-1} \times 0.1 \ m = 0.186 \ K$ (or $0.186^{\circ}C$).
Freezing point of solution = Freezing point of pure solvent - $\Delta T_f$.
Freezing point of solution = $0^{\circ}C - 0.186^{\circ}C = -0.186^{\circ}C$.

(C) The depression in freezing point is given by the formula: $\Delta T_f = i \times K_f \times m$.
Since the molality $(m = 0.01 \ mol \ kg^{-1})$ is constant and the van't Hoff factor $(i)$ is $1$ for non-electrolytic solutes in these solvents,the depression in freezing point is directly proportional to the cryoscopic constant $(K_f)$: $\Delta T_f \propto K_f$.
Comparing the given $K_f$ values:
Water: $1.86$
Benzene: $5.12$
Carbon tetrachloride: $31.8$
Cyclohexane: $20.0$
Since carbon tetrachloride has the highest $K_f$ value $(31.8)$,the depression in freezing point will be the highest for this solvent.

(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: $w_1 = 2 \ g$,$m_1 = 500 \ g \ mol^{-1}$,$w_2 = 57.3 \ g$,$K_f = 4.3 \ K \ kg \ mol^{-1}$.
Substituting the values: $\Delta T_f = 4.3 \times \frac{2 \times 1000}{500 \times 57.3}$.
$\Delta T_f = 4.3 \times \frac{2000}{28650} = 4.3 \times 0.0698 \approx 0.3 \ K$.

(C) 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.
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 $g$.
Molar mass of ascorbic acid $(C_6H_8O_6) = (6 \times 12) + (8 \times 1) + (6 \times 16) = 72 + 8 + 96 = 176 \ g \ mol^{-1}$.
Given: $\Delta T_f = 1.5 \ K$,$K_f = 3.9 \ K \ kg \ mol^{-1}$,$w_1 = 100 \ g$.
Substituting the values: $1.5 = \frac{3.9 \times w_2 \times 1000}{176 \times 100}$.
$1.5 = \frac{39 \times w_2}{176}$.
$w_2 = \frac{1.5 \times 176}{39} = \frac{264}{39} \approx 6.77 \ g$.
Rounding to the nearest provided option,the correct value is $6.6 \ g$.

(B) Given: Mass of solute $= 36 \ g$,Mass of solvent $= 1.2 \ kg$,$\Delta T_f = 0.93 \ ^\circ C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Using the formula $\Delta T_f = K_f \times m$,where $m = \frac{w_2}{M_2 \times w_1(kg)}$.
Substituting the values: $0.93 = \frac{1.86 \times 36}{M_2 \times 1.2}$.
Solving for $M_2$: $M_2 = \frac{1.86 \times 36}{0.93 \times 1.2} = 60 \ g \ mol^{-1}$.
Empirical formula mass of $CH_2O = 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2$.
Molecular formula $= n \times (CH_2O) = 2 \times CH_2O = C_2H_4O_2$.

(B) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given $\Delta T_f = 0 - (-0.93) = 0.93 \ K$.
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
For a $7 \%$ by mass solution,$7 \ g$ of solute is present in $93 \ g$ of solvent (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: $0.93 = 1.86 \times \frac{7}{M \times 0.093}$.
$M = \frac{1.86 \times 7}{0.93 \times 0.093} = \frac{2 \times 7}{0.093} = \frac{14}{0.093} \approx 150.53 \ g \ mol^{-1}$.
Thus,the molar mass is approximately $150.5 \ g \ mol^{-1}$.

(B) Given:
$W_B$ (mass of ethylene glycol) = $31 \ g$
$W_A$ (mass of water) = $600 \ g$
$K_f$ (for water) = $1.86 \ K \ kg \ mol^{-1}$
$M_B$ (molar mass of $C_2H_6O_2$) = $(2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g \ mol^{-1}$
Formula for freezing point depression: $\Delta T_f = K_f \times \frac{W_B}{M_B} \times \frac{1000}{W_A(g)}$
Substituting the values:
$\Delta T_f = \frac{1.86 \times 31 \times 1000}{62 \times 600}$
$\Delta T_f = \frac{1.86 \times 31000}{37200} = \frac{57660}{37200} = 1.55 \ K$

(D) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Where $\Delta T_f = T_f^{\circ} - T_f = 0 - (-0.93) = 0.93 \ K$.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2 = 6 \ g$,$w_1 = 100 \ g$,and $M_2$ is the molar mass of $X$.
Substituting the values: $0.93 = 1.86 \times \frac{6 \times 1000}{M_2 \times 100}$.
$0.93 = \frac{1.86 \times 60}{M_2}$.
$M_2 = \frac{1.86 \times 60}{0.93} = 2 \times 60 = 120 \ g \ mol^{-1}$.

(A) The formula for the molal depression constant is $K_{f} = \frac{R T_{f}^2 M_{solvent}}{1000 \times L_{f}}$,where $L_{f}$ is the latent heat of fusion in $J \ g^{-1}$.
Given: $T_{f} = 17 + 273 = 290 \ K$,$L_{f} = 180 \ J \ g^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Assuming the molar mass of the solvent $(M_{solvent})$ is $1 \ g \ mol^{-1}$ as per the context of the calculation provided:
$K_{f} = \frac{8.314 \times (290)^2 \times 1}{1000 \times 180}$
$K_{f} = \frac{8.314 \times 84100}{180000} = \frac{699207.4}{180000} \approx 3.88 \ K \ kg \ mol^{-1}$.