Depression of freezing point of the solvent Questions in English

(D) Camphor is used in the Rast method for the determination of molecular mass of organic compounds.
This is because camphor has a very high cryoscopic constant $(K_f = 37.7 \ K \ kg \ mol^{-1})$,which results in a large depression in freezing point even for small amounts of solute,making the measurement accurate.

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

(D) The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality.
Molality $m = \frac{w \times 1000}{M \times W}$,where $w$ is the mass of solute (urea,$M = 60 \ g/mol$) and $W$ is the mass of solvent ($500 \ g$ of water).
Given $\Delta T_f = 0.186 \ ^oC$,$K_f = 18.6 \ K \ kg \ mol^{-1}$,and $W = 500 \ g$.
Substituting the values: $0.186 = 18.6 \times \frac{w \times 1000}{60 \times 500}$.
$0.186 = 18.6 \times \frac{w}{30}$.
$0.186 = 0.62 \times w$.
$w = \frac{0.186}{0.62} = 0.3 \ g$.

(A) The freezing point depression constant $(K_f)$ is a characteristic property of a solvent. Among the given options,Camphor has the highest $K_f$ value,which is approximately $39.7 \ K \ kg \ mol^{-1}$. Therefore,the correct option is $(A)$.

(A) The freezing point of pure water is $T_f^0 = 273.15 \ K$.
Given the freezing point of the solution is $T_f = 271.9 \ K$.
The depression in freezing point is $\Delta T_f = T_f^0 - T_f = 273.15 \ K - 271.9 \ K = 1.25 \ K$.
The formula for molar mass $M_2$ is: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$.
Substituting the values: $M_2 = \frac{1.86 \times 1.25 \times 1000}{1.25 \times 20}$.
$M_2 = \frac{1860}{20} = 93 \ g \ mol^{-1}$.
Wait,re-calculating with $\Delta T_f = 273.0 - 271.9 = 1.1 \ K$:
$M_2 = \frac{1.86 \times 1.25 \times 1000}{1.1 \times 20} = \frac{2325}{22} \approx 105.68 \ g \ mol^{-1}$.
Thus,the molar mass is approximately $105.7 \ g \ mol^{-1}$.

(B) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$.
Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$ and molality $m = 0.1 \ m$.
Substituting the values: $\Delta T_f = 1.86 \times 0.1 = 0.186 \ K$.

(B) The freezing point of a pure solvent is lowered upon the addition of a non-volatile solute,a phenomenon known as depression of freezing point.
Since $Ca(NO_3)_2$ is a solute,the freezing point of its $1\%$ solution will be lower than the freezing point of pure water $(0\ ^oC)$.
Therefore,the freezing point is less than $0\ ^oC$.

(A) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $342\,g\,mol^{-1}$.
Given mass of solute $(w_2)$ = $342\,g$,mass of solvent $(w_1)$ = $1000\,g$,and $K_f = 1.86\,^{\circ}C\,kg\,mol^{-1}$.
Molality $(m)$ = $\frac{w_2 \times 1000}{M_2 \times w_1} = \frac{342 \times 1000}{342 \times 1000} = 1\,mol\,kg^{-1}$.
Depression in freezing point $(\Delta T_f)$ = $K_f \times m = 1.86 \times 1 = 1.86\,^{\circ}C$.
Freezing point of solution $(T_f)$ = $T_f^{\circ} - \Delta T_f = 0\,^{\circ}C - 1.86\,^{\circ}C = - 1.86\,^{\circ}C$.

(D) 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 molal freezing point depression constant (cryoscopic constant) and $m$ is the molality of the solution.
Rearranging the equation,we get: $\frac{\Delta T_f}{m} = K_f$.
Since $K_f$ is a constant characteristic of the solvent,the limit as $m \to 0$ of the ratio $\frac{\Delta T_f}{m}$ is simply $K_f$.
Therefore,the quantity $\lim_{m \to 0} \left( \frac{\Delta T_f}{m} \right)$ is equal to $K_f$.

(A) The freezing point of a pure solvent is lowered by the addition of a non-volatile solute,a phenomenon known as depression of freezing point.
Since lead nitrate $(Pb(NO_3)_2)$ is a solute,its addition to water will cause the freezing point of the solution to decrease below the freezing point of pure water $(0 \, ^\circ C)$.
Therefore,the freezing point of the $1 \%$ solution will be below $0 \, ^\circ C$.

(A) During the depression of freezing point in a solution,the liquid solvent and the solid solvent are in equilibrium.
During the freezing process of a solution,only the solvent molecules transition into the solid phase.
At the freezing point,the vapour pressure of the solid solvent and the liquid solvent must be equal.
If the vapour pressures were not equal,the system would not be at equilibrium.
The presence of a solute lowers the vapour pressure of the liquid solvent,which consequently lowers the temperature at which the liquid and solid forms of the solvent reach equilibrium.

(A) The formula for the molar mass $(M)$ of a solute using the depression in freezing point is:
$M = \frac{K_f \times w \times 1000}{\Delta T_f \times W_{solvent}}$
Given:
$K_f = 5.12 \ K \ kg \ mol^{-1}$
$w = 1.00 \ g$
$W_{solvent} = 50 \ g$
$\Delta T_f = 0.40 \ K$
Substituting the values:
$M = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50} = \frac{5120}{20} = 256 \ g \ mol^{-1}$
Thus,the molecular mass of the solute is $256 \ g \ mol^{-1}$.

(A) The formula for molar mass $(M_2)$ using depression in freezing point is:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Given:
$K_f = 5.12 \ ^oC \ kg \ mol^{-1}$
$w_2 = 0.440 \ g$
$w_1 = 22.2 \ g$
$\Delta T_f = 0.567 \ ^oC$
Substituting the values:
$M_2 = \frac{5.12 \times 0.440 \times 1000}{0.567 \times 22.2} = \frac{2252.8}{12.5874} \approx 178.9 \ g \ mol^{-1}$
Thus,the correct option is $A$.

(A) The depression in freezing point is given by the formula $\Delta T_f = i \times K_f \times m$.
Since $K_f$ and $m$ are constant for all solutions,the freezing point depends on the van't Hoff factor $(i)$.
Lower $i$ value results in a smaller depression in freezing point,which means a higher freezing point.
$1$. Urea $(NH_2CONH_2)$ is a non-electrolyte,so $i = 1$.
$2$. Potassium bromide $(KBr)$ dissociates as $KBr \rightarrow K^+ + Br^-$,so $i = 2$.
$3$. Barium chloride $(BaCl_2)$ dissociates as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,so $i = 3$.
$4$. Aluminium sulphate $(Al_2(SO_4)_3)$ dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so $i = 5$.
Since Urea has the lowest van't Hoff factor $(i = 1)$,it will have the highest freezing point.

(D) The depression in freezing point is a colligative property,which depends on the number of particles (ions) produced in the solution.
$1$. $NaCl \to Na^{+} + Cl^{-}$ ($2$ ions)
$2$. Urea is a non-electrolyte ($1$ particle)
$3$. Glucose is a non-electrolyte ($1$ particle)
$4$. $K_2SO_4 \to 2K^{+} + SO_4^{2-}$ ($3$ ions)
Since $K_2SO_4$ produces the maximum number of ions ($3$ ions) per formula unit,it will exhibit the maximum depression in freezing point.

(B) The depression in freezing point $\Delta T_f$ is directly proportional to the molarity $M$ of the solution,i.e.,$\Delta T_f \propto M$.
Given,for $M_1 = 0.01 \ M$,$\Delta T_{f,1} = 0.18 \ ^oC$ (since freezing point is $-0.18 \ ^oC$).
When equal volumes $(V)$ of two solutions are mixed,the final concentration $M_{mix}$ is given by $M_{mix} = \frac{M_1 V + M_2 V}{V + V} = \frac{M_1 + M_2}{2}$.
$M_{mix} = \frac{0.01 + 0.002}{2} = 0.006 \ M$.
Using the proportionality $\frac{\Delta T_{f,2}}{\Delta T_{f,1}} = \frac{M_{mix}}{M_1}$,we get $\Delta T_{f,2} = \frac{0.006}{0.01} \times 0.18 = 0.6 \times 0.18 = 0.108 \ ^oC$.
Thus,the new freezing point is $-0.108 \ ^oC$,which is approximately $-0.10 \ ^oC$.

(A) The molar mass of $C_2H_5OH$ is $(2 \times 12) + (6 \times 1) + 16 = 46 \ g \ mol^{-1}$.
The molality $(m)$ of the solution is calculated as: $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{17 \ g}{46 \ g \ mol^{-1} \times 1 \ kg} = 0.3696 \ mol \ kg^{-1}$.
The depression in freezing point is given by: $\Delta T_f = K_f \times m = 1.86 \ K \ kg \ mol^{-1} \times 0.3696 \ mol \ kg^{-1} \approx 0.6875 \ ^oC \approx 0.69 \ ^oC$.
The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0 \ ^oC - 0.69 \ ^oC = - 0.69 \ ^oC$.

(B) The formula for the molal depression constant $(K_f)$ is given by: $K_f = \frac{R \times T_f^2}{1000 \times L_f}$
Given values:
$R = 8.314\,J\,K^{-1}\,mol^{-1}$
$T_f = 273 + 16.6 = 289.6\,K$
$L_f = 180.75\,J\,g^{-1}$
Substituting the values into the formula:
$K_f = \frac{8.314 \times (289.6)^2}{1000 \times 180.75}$
$K_f = \frac{8.314 \times 83868.16}{180750}$
$K_f = \frac{697311.66}{180750} \approx 3.86\,K\,kg\,mol^{-1}$
Thus,the correct option is $B$.

(B) The depression in freezing point $(\Delta T_f)$ is a colligative property,which depends on the number of solute particles in the solution.
The relationship is given by $\Delta T_f = i \times K_f \times m$.
Since the freezing point $T_f = T_f^0 - \Delta T_f$,a smaller value of $\Delta T_f$ results in a higher freezing point.
Sugar is a non-electrolyte $(i = 1)$,while $NaCl$ $(i = 2)$,$BaCl_2$ $(i = 3)$,and $FeCl_3$ $(i = 4)$ dissociate into ions.
Among the given options,the sugar solution has the minimum number of particles,leading to the minimum depression in freezing point and thus the highest freezing point.

(C) The depression in freezing point is a colligative property,which depends on the van't Hoff factor $(i)$.
For $0.1 \ M$ glucose,$i = 1$.
For $0.1 \ M$ sodium chloride $(NaCl)$,$i = 2$.
For $0.1 \ M$ barium chloride $(BaCl_2)$,$i = 3$.
For $0.1 \ M$ magnesium sulphate $(MgSO_4)$,$i = 2$.
Since $BaCl_2$ provides the maximum number of ions $(i = 3)$,it will produce the maximum depression in freezing point.

(D) The depression in freezing point $(\Delta T_f)$ is a colligative property,which is directly proportional to the van't Hoff factor $(i)$.
$(i)$ For $K_2Cr_2O_7$,$i = 3$ $(2K^+ + Cr_2O_7^{2-})$.
$(ii)$ For $NH_4Cl$,$i = 2$ $(NH_4^+ + Cl^-)$.
$(iii)$ For $BaSO_4$,$i = 2$ $(Ba^{2+} + SO_4^{2-})$.
$(iv)$ For $Al_2(SO_4)_3$,$i = 5$ $(2Al^{3+} + 3SO_4^{2-})$.
Since $Al_2(SO_4)_3$ produces the maximum number of particles $(i = 5)$,it will show the maximum depression in freezing point,resulting in the minimum freezing point.

(B) The depression in freezing point is $\Delta T_f = T_f^\circ - T_f$.
Taking the freezing point of pure water $T_f^\circ = 273 \text{ K}$,we have:
$\Delta T_f = 273 - 272.813 = 0.187 \text{ K}$.
The molecular mass $(M)$ is calculated using the formula:
$M = \frac{K_f \times w \times 1000}{\Delta T_f \times W}$
Substituting the given values:
$M = \frac{1.86 \times 0.1 \times 1000}{0.187 \times 21.7} = \frac{186}{4.0579} \approx 45.83 \text{ g/mol}$.
The molecular mass is approximately $46 \text{ g/mol}$.

(A) The depression in freezing point $(\Delta T_f)$ is calculated using the formula: $\Delta T_f = K_f \times m$.
Given: $K_f = 4.9 \, K \, kg \, mol^{-1}$ and molality $(m) = 0.001 \, m$.
Substituting the values: $\Delta T_f = 4.9 \times 0.001 = 0.0049 \, K$.
Therefore,the correct option is $A$.

(C) Glycerine is a non-volatile solute. When a non-volatile solute is added to a solvent,it leads to the depression of the freezing point of the solvent. This is a colligative property known as freezing point depression,represented by the formula $\Delta T_f = K_f \times m$,where $\Delta T_f$ is the depression in freezing point,$K_f$ is the cryoscopic constant,and $m$ is the molality of the solution.

(A) Ethylene glycol is added to water as an antifreeze agent. It lowers the freezing point of water,preventing it from freezing in the radiators during cold weather.

(B) The freezing point is the temperature at which the liquid and solid forms of the same substance are in equilibrium and have the same vapour pressure.
When a solute is dissolved in a solvent,the vapour pressure of the solution decreases,leading to a depression in the freezing point.
In an aqueous solution of glucose,the solute (glucose) is non-volatile,and the solvent is water.
Upon cooling,the solvent (water) reaches its freezing point first and begins to crystallize out as pure solid ice.
Therefore,crystals of water will be separated out first.

(D) Camphor is used in the determination of molecular mass by the Rast method because it has a very high cryoscopic constant $(K_f = 37.7 \ K \ kg \ mol^{-1})$. This allows for a large depression in the freezing point,which can be measured accurately even with small amounts of the solute.

(A) The molality $m$ of the solution is calculated as: $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}}$.
Given mass of $C_2H_5OH = 17 \ g$,molar mass of $C_2H_5OH = 46 \ g/mol$,and mass of water $= 1000 \ g = 1 \ kg$.
$m = \frac{17}{46 \times 1} = 0.3696 \ mol/kg$.
The depression in freezing point is $\Delta T_f = K_f \times m = 1.86 \times 0.3696 = 0.6875 \ ^oC \approx 0.69 \ ^oC$.
Since $\Delta T_f = T_f^o - T_f$,where $T_f^o = 0 \ ^oC$ for water:
$0.69 = 0 - T_f \Rightarrow T_f = -0.69 \ ^oC$.

(B) The freezing point depression is given by $\Delta T_f = i \times K_f \times m$.
Since the molality $(m)$ and $K_f$ are constant for all solutions,the freezing point depression depends on the van't Hoff factor $(i)$.
Lower $i$ value leads to lower freezing point depression,which results in a higher freezing point.
For $0.1 \ M \ NaCl$,$i = 2$.
For $0.1 \ M \ Sugar$,$i = 1$.
For $0.1 \ M \ BaCl_2$,$i = 3$.
For $0.1 \ M \ FeCl_3$,$i = 4$.
Since $Sugar$ has the lowest van't Hoff factor $(i = 1)$,it will have the minimum freezing point depression and thus the highest freezing point.

(D) The depression of freezing point occurs because the vapor pressure of the solution is lower than that of the pure solvent at the same temperature.
At the freezing point,the solid phase and the liquid phase are in equilibrium.
Since the vapor pressure of the solution is lower,it reaches the vapor pressure of the solid solvent at a lower temperature.
During this process,only the solvent molecules solidify,while the solute molecules remain in the liquid phase.

(A) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Where $\Delta T_f$ is the depression in freezing point,$K_f$ is the molal freezing point depression constant,and $m$ is the molality.
Given: $\Delta T_f = 0.184 \ K$ and $K_f = 18.4 \ K \ kg \ mol^{-1}$.
Substituting the values: $m = \frac{\Delta T_f}{K_f} = \frac{0.184}{18.4} = 0.01 \ mol \ kg^{-1}$.

(B) The freezing point of a solvent decreases when a non-volatile solute is added to it,a phenomenon known as depression of freezing point.
Since $Ca(NO_3)_2$ is a solute,the freezing point of its aqueous solution will be lower than that of pure water $(0\,^oC)$.
Therefore,the freezing point will be less than $0\,^oC$.

(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 mass of solute is $5 \ g$ and the mass of solvent is $95 \ g$.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$.
Since $w_2$ and $w_1$ are the same for both cane sugar $(M_2 = 342 \ g/mol)$ and glucose $(M_2 = 180 \ g/mol)$,$\Delta T_f$ is inversely proportional to the molar mass $M_2$.
$\Delta T_{f, \text{sugar}} = 273.15 - 271 = 2.15 \ K$.
$\frac{\Delta T_{f, \text{glucose}}}{\Delta T_{f, \text{sugar}}} = \frac{M_{\text{sugar}}}{M_{\text{glucose}}} = \frac{342}{180} = 1.9$.
$\Delta T_{f, \text{glucose}} = 1.9 \times 2.15 = 4.085 \ K$.
Freezing point of glucose solution $= 273.15 - 4.085 = 269.065 \ K \approx 269.07 \ K$.

(B) The formula for freezing point depression 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) and $W_1$ is the mass of solvent (water).
Given: $\Delta T_f = 0.186^oC$,$K_f = 1.86^oC \ kg \ mol^{-1}$,$W_1 = 500 \ g$,Molar mass of urea $(NH_2CONH_2)$ $M_2 = 60 \ g \ mol^{-1}$.
Substituting the values: $0.186 = 1.86 \times \frac{W_2 \times 1000}{60 \times 500}$.
$0.186 = 1.86 \times \frac{W_2 \times 2}{60}$.
$0.186 = 1.86 \times \frac{W_2}{30}$.
$W_2 = \frac{0.186 \times 30}{1.86} = 0.1 \times 30 = 3 \ g$.
Therefore,$3 \ g$ of urea is required.

(D) Given:
Mass of solute $(W_1)$ = $4.5 \ g$
Mass of solvent $(W_2)$ = $100 \ g$
Depression in freezing point $(\Delta T_f)$ = $0.465 \ K$
Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$
The formula for molar mass $(M_1)$ is:
$M_1 = \frac{K_f \times 1000 \times W_1}{\Delta T_f \times W_2}$
Substituting the values:
$M_1 = \frac{1.86 \times 1000 \times 4.5}{0.465 \times 100}$
$M_1 = \frac{1.86 \times 45}{0.465}$
$M_1 = 180 \ g/mol$

(B) Given: $(T_f)_s = 271.9 \ K$,$w = 1.25 \ g$,$W = 20 \ g$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$T_0 = 273 \ K$.
$\Delta T_f = T_0 - (T_f)_s = 273 - 271.9 = 1.1 \ K$.
The formula for depression in freezing point is:
$\Delta T_f = \frac{w \times 1000 \times K_f}{M_w \times W}$.
Rearranging for molar mass $(M_w)$:
$M_w = \frac{w \times 1000 \times K_f}{\Delta T_f \times W}$.
Substituting the values:
$M_w = \frac{1.25 \times 1000 \times 1.86}{1.1 \times 20}$.
$M_w = \frac{2325}{22} = 105.68 \ g \ mol^{-1}$.

(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.
Here,$\Delta T_f = 0 - (-6) = 6 \ K$.
$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Mass of solvent $(W_A)$ $= 4 \ kg$.
Molar mass of solute $(M_B)$ $= 62 \ g \ mol^{-1}$.
Molality $(m)$ $= \frac{W_B}{M_B \times W_A (\text{in } kg)}$.
Substituting the values: $6 = 1.86 \times \frac{W_B}{62 \times 4}$.
$W_B = \frac{6 \times 62 \times 4}{1.86}$.
$W_B = \frac{1488}{1.86} = 800 \ g$.

(B) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Given,$K_f = 1.86$ and molality $m = 0.1 \ m$.
Substituting the values: $\Delta T_f = 1.86 \times 0.1 = 0.186 \ K$.

(C) The depression in freezing point is given by $\Delta T_f = K_f \times m$,where $m$ is the molality.
For $5\%$ (by mass) solutions,the mass of solute is $5 \, g$ and the mass of solvent is $95 \, g$.
$\Delta T_f = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$.
Since $w_2$,$w_1$,and $K_f$ are the same for both,$\Delta T_f \propto \frac{1}{M_2}$.
$\Delta T_f(\text{sucrose}) = 273.15 - 271 = 2.15 \, K$.
$\frac{\Delta T_f(\text{glucose})}{\Delta T_f(\text{sucrose})} = \frac{M(\text{sucrose})}{M(\text{glucose})} = \frac{342}{180} = 1.9$.
$\Delta T_f(\text{glucose}) = 2.15 \times 1.9 = 4.085 \, K$.
Freezing point of glucose solution $= 273.15 - 4.085 = 269.065 \, K \approx 269.07 \, K$.

(D) The depression of freezing point is given by the formula: $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Rearranging this,we get $\frac{\Delta T_f}{K_f} = m$.
Given that $\frac{\Delta T_f}{K_f} = \frac{1}{1000} = 0.001 \ mol/kg$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Since the density of water is $1 \ g/mL$,$1 \ L$ of water is $1 \ kg$.
Therefore,the number of moles of glucose required is $0.001 \ mol$.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.
Mass of glucose = $\text{moles} \times \text{molar mass} = 0.001 \ mol \times 180 \ g/mol = 0.18 \ g$.

(A) The freezing point of pure water $(T_f^0)$ is $273.15 \ K$.
The depression in freezing point is $\Delta T_f = T_f^0 - T_f = 273.15 \ K - 271.9 \ K = 1.25 \ K$.
The formula for molar mass $(M_2)$ of the solute is $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$.
Substituting the given values: $M_2 = \frac{1.86 \times 1.25 \times 1000}{1.25 \times 20}$.
$M_2 = \frac{1.86 \times 1000}{20} = \frac{1860}{20} = 93 \ g \ mol^{-1}$.
Note: Based on the provided options,the calculation $1.86 \times 1.25 \times 1000 / (1.25 \times 20)$ yields $93$. However,if $\Delta T_f$ was intended to be $1.1 \ K$ as per the provided solution snippet,the result is $105.68 \approx 105.7$.

(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$,where $m$ is the molality of the solution.
Given: $\Delta T_f = 2.8 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,mass of solvent $(W_{solvent}) = 1.0 \ kg$,and molar mass of ethylene glycol $(C_2H_6O_2) = 62 \ g \ mol^{-1}$.
Substituting the values: $2.8 = 1.86 \times \frac{W_{solute}}{62 \times 1.0}$.
Solving for $W_{solute}$: $W_{solute} = \frac{2.8 \times 62}{1.86} \approx 93.33 \ g$.
Thus,the required amount is approximately $93 \ g$.

(A) The depression in freezing point $\Delta T_f$ is directly proportional to the van't Hoff factor $(i)$ and molar concentration $(C)$,i.e.,$\Delta T_f \propto C \times i$.
For $0.01 \, M \ NaCl$: $C \times i = 0.01 \times 2 = 0.020$.
For $0.005 \, M \ C_2H_5OH$: $C \times i = 0.005 \times 1 = 0.005$.
For $0.005 \, M \ MgI_2$: $C \times i = 0.005 \times 3 = 0.015$.
For $0.005 \, M \ MgSO_4$: $C \times i = 0.005 \times 2 = 0.010$.
Since the product $C \times i$ is highest for $0.01 \, M \ NaCl$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

(A) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Given:
$K_f = 4.9 \ K \ kg \ mol^{-1}$
$m = 0.001 \ m$
Substituting the values:
$\Delta T_f = 4.9 \times 0.001 = 0.0049 \ K$

(C) For cane sugar solution: $(\Delta T_f)_{cane} = K_f \times m = K_f \times \frac{5 \times 1000}{342 \times 100} = 273.15 - 271 = 2.15 \, K$
For glucose solution: $(\Delta T_f)_{glucose} = K_f \times m = K_f \times \frac{5 \times 1000}{180 \times 100}$
Dividing the two equations: $\frac{(\Delta T_f)_{glucose}}{2.15} = \frac{342}{180}$
$(\Delta T_f)_{glucose} = 2.15 \times \frac{342}{180} = 4.085 \, K$
Freezing point of glucose solution = $273.15 - 4.085 = 269.065 \, K \approx 269.07 \, K$.

(B) The formula for the molal freezing point depression constant is given by: $K_f = \frac{R \times T_f^2}{1000 \times L_f}$
Given values:
$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$
$T_f = 273.15 + 16.6 = 289.75 \, K$
$L_f = 180.75 \, J/g$
Substituting the values:
$K_f = \frac{8.314 \times (289.75)^2}{1000 \times 180.75}$
$K_f = \frac{8.314 \times 83955.06}{180750}$
$K_f = \frac{698022.37}{180750} \approx 3.86 \, K \cdot kg \cdot mol^{-1}$

(B) For the first solution: $\Delta T_f = 0 - (-0.18) = 0.18^\circ C$.
Since $\Delta T_f = K_f \times m$,we have $0.18 = K_f \times 0.01$,so $K_f = 18$.
When equal volumes of two solutions are mixed,the new molality $m_{mix}$ is the average of the two molalities: $m_{mix} = \frac{m_1 + m_2}{2} = \frac{0.01 + 0.002}{2} = 0.006 \ m$.
The new depression in freezing point is $\Delta T_{f(new)} = K_f \times m_{mix} = 18 \times 0.006 = 0.108^\circ C$.
Therefore,the new freezing point is $0 - 0.108 = -0.108^\circ C$.

(B) Statement $1$ is true because at the freezing point,the vapor pressure of the solid phase equals the vapor pressure of the liquid phase,leading to crystallization.
Statement $2$ is true because the depression of freezing point $(\Delta T_f)$ is defined as $T_f^0 - T_f$,where $T_f^0$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution.
However,Statement $2$ describes the definition of the property,not the physical reason why crystallization occurs at that specific temperature. Thus,Statement $2$ is not the correct explanation for Statement $1$.

(B) The freezing point of a solution is always lower than the freezing point of the pure solvent due to the depression of freezing point phenomenon.
Since $Ca(NO_3)_2$ is a solute,the freezing point of its aqueous solution will be lower than that of pure water,which is $0^\circ C$.

(B) Let $a$ and $b$ be the atomic weights of $A$ and $B$ respectively.
For $AB_2$,molar mass $M_1 = a + 2b$.
For $AB_4$,molar mass $M_2 = a + 4b$.
Using the formula $\Delta T_f = \frac{1000 \times K_f \times w}{M \times W}$:
For $AB_2$: $2.3 = \frac{1000 \times 5.1 \times 1}{(a + 2b) \times 20} \implies a + 2b = \frac{5100}{2.3 \times 20} \approx 110.87 \dots (1)$
For $AB_4$: $1.3 = \frac{1000 \times 5.1 \times 1}{(a + 4b) \times 20} \implies a + 4b = \frac{5100}{1.3 \times 20} \approx 196.15 \dots (2)$
Subtracting $(1)$ from $(2)$: $(a + 4b) - (a + 2b) = 196.15 - 110.87 \implies 2b = 85.28 \implies b = 42.64$.
Substituting $b$ in $(1)$: $a + 2(42.64) = 110.87 \implies a + 85.28 = 110.87 \implies a = 25.59$.
Thus,the atomic weights are $A = 25.59$ and $B = 42.64$.