Depression of freezing point of the solvent Questions in English

(D) The depression in freezing point is given by the formula: $\Delta T_f = K_f \times m$
Here,$m$ is the molality of the solution.
The molar mass of ethanol $(C_2H_5OH)$ is $M_B = (2 \times 12) + (6 \times 1) + 16 = 46 \ g \ mol^{-1}$.
The mass of solute $(w_B)$ is $25 \ g$ and the mass of solvent $(w_A)$ is $1000 \ g$.
$\Delta T_f = K_f \times \frac{w_B \times 1000}{M_B \times w_A} = 1.86 \times \frac{25 \times 1000}{46 \times 1000} = 1.86 \times 0.543 \approx 1.01^{\circ} C$.
The freezing point of the solution is $T_f = T_f^{\circ} - \Delta T_f = 0^{\circ} C - 1.01^{\circ} C = -1.01^{\circ} C$.
Rounding to the nearest integer,the freezing point is $-1^{\circ} C$.

(A) The formula for the molal depression constant $(K_f)$ is given by:
$K_f = \frac{R \times T_f^2}{1000 \times L_f}$
Where:
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$T_f = 273 + 15 = 288 \ K$
$L_f = 180.7 \ J \ g^{-1}$
Substituting the values:
$K_f = \frac{8.314 \times (288)^2}{1000 \times 180.7}$
$K_f = \frac{8.314 \times 82944}{180700}$
$K_f = \frac{689624.416}{180700} \approx 3.81 \ K \ kg \ mol^{-1}$

(D) Given:
Volume of solute $A = 10 \ mL$
$\Delta T_f = 0 - (-0.413) = 0.413 \ K$
Mass of solvent $(w_A) = 500 \ g$
$K_f \text{ (water)} = 1.86 \ K \ kg \ mol^{-1}$
Molecular weight of $A \ (M_B) = 60 \ g \ mol^{-1}$
Using the formula for freezing point depression:
$\Delta T_f = K_f \times \frac{w_B}{M_B} \times \frac{1000}{w_A}$
$0.413 = 1.86 \times \frac{w_B}{60} \times \frac{1000}{500}$
$\therefore w_B = \frac{0.413 \times 60 \times 500}{1.86 \times 1000} = 6.66 \ g$
Total mass of solution $= w_A + w_B = 500 + 6.66 = 506.66 \ g$
Total volume of solution $= V_{\text{solvent}} + V_{\text{solute}} = 500 \ mL + 10 \ mL = 510 \ mL$ (Assuming density of water $\approx 1 \ g \ mL^{-1}$)
Density of solution $(d) = \frac{\text{Total mass}}{\text{Total volume}} = \frac{506.66 \ g}{510 \ mL} \approx 0.993 \ g \ mL^{-1}$
Thus,the correct option is $(D)$.

(C) Melting point of pure camphor $= 176^{\circ} C$
$K_f$ for camphor $= 40 \ K \ kg \ mol^{-1}$
Mass of hydrocarbon solute $(w_B) = 0.02 \ g$
Mass of camphor solvent $(w_A) = 0.8 \ g$
Freezing point of solution $= 156.77^{\circ} C$
Depression in freezing point,$\Delta T_f = 176 - 156.77 = 19.23 \ K$
Using the formula $\Delta T_f = \frac{K_f \times w_B \times 1000}{M_B \times w_A}$:
$19.23 = \frac{40 \times 0.02 \times 1000}{M_B \times 0.8}$
$M_B = \frac{40 \times 0.02 \times 1000}{19.23 \times 0.8} = \frac{800}{15.384} \approx 52 \ g \ mol^{-1}$
For the hydrocarbon with $92.3 \%$ carbon,the remaining $7.7 \%$ is hydrogen.
Empirical formula calculation:
$C: \frac{92.3}{12} = 7.69 \approx 7.7$
$H: \frac{7.7}{1} = 7.7$
Ratio $C:H = 1:1$,so empirical formula is $CH$ (empirical mass $= 13 \ g \ mol^{-1}$).
$n = \frac{\text{Molar mass}}{\text{Empirical mass}} = \frac{52}{13} = 4$
Molecular formula $= (CH)_4 = C_4 H_4$.

(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,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent.
Substituting the given values: $1.2 = 5.12 \times \frac{2.0 \times 1000}{M_2 \times 100}$.
$1.2 = \frac{5.12 \times 20}{M_2}$.
$M_2 = \frac{102.4}{1.2} \approx 85.33 \ g \ mol^{-1}$.
Rounding to the nearest integer,the molar mass is $85 \ g \ mol^{-1}$.

(A) The freezing point of a substance is defined as the temperature at which the solid and liquid phases of the substance are in equilibrium.
In the context of the depression in freezing point experiment,the equilibrium is established between the molecules of the liquid solvent (present in the solution) and the solid solvent (which separates out upon freezing).

(C) The depression in freezing point is given by $\Delta T_{f} = K_{f} \times m$.
Here,$10 \%$ mass of ethylene glycol ($C_2H_6O_2$,molar mass $= 62 \ g \ mol^{-1}$) means $10 \ g$ of solute in $90 \ g$ of water.
Molality $(m)$ $= \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{10}{62} \times \frac{1000}{90} \approx 1.79 \ mol \ kg^{-1}$.
$\Delta T_{f} = 1.86 \times 1.79 \approx 3.33^{\circ} C$.
Since the freezing point of pure water is $0^{\circ} C$,the temperature at which ice begins to separate is $0 - 3.33 = -3.33^{\circ} C$,which is approximately $-3.3^{\circ} C$.

(C) First, calculate the molar mass of ascorbic acid $(\text{C}_6\text{H}_8\text{O}_6)$:
$\text{Molar mass} = (6 \times 12) + (8 \times 1) + (6 \times 16) = 72 + 8 + 96 = 176 \text{ g/mol}$.
Use the formula for depression in freezing point: $\Delta T_f = K_f \times m$, where $m$ is the molality.
$m = \frac{w \times 1000}{M \times W_{\text{solvent}}}$, where $w$ is the mass of solute and $W_{\text{solvent}}$ is the mass of solvent in grams.
Substituting the given values: $1.5 = 3.9 \times \frac{w \times 1000}{176 \times 75}$.
Solving for $w$: $w = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = \frac{19800}{3900} \approx 5.077 \text{ g}$.