Abnormal molecular mass Questions in English

(D) The formula for elevation of boiling point is $\Delta T_{b} = i \times K_{b} \times m$.
Given $\Delta T_{b} = 1.518 \ K$,$K_{b} = 2.53 \ K \ kg \ mol^{-1}$,and $m = 1 \ mol \ kg^{-1}$.
For dimerization,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
Substituting the values: $1.518 = (1 - \frac{\alpha}{2}) \times 2.53 \times 1$.
$\frac{1.518}{2.53} = 1 - \frac{\alpha}{2}$.
$0.6 = 1 - \frac{\alpha}{2}$.
$\frac{\alpha}{2} = 1 - 0.6 = 0.4$.
$\alpha = 0.8$,which is $80 \%$.

(B) Given: $T_f^o = 5.3\,^oC$,$T_f = 4.47\,^oC$,$\Delta T_f = 5.3 - 4.47 = 0.83\, K$.
Mass of solute $(w_2) = 0.223\, g$,Mass of solvent $(w_1) = 4.4\, g = 0.0044\, kg$.
Molar mass of $C_6H_5CH_2COOH$ $(M_2) = 136\, g/mol$.
Using the formula: $\Delta T_f = i \times K_f \times m = i \times K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$.
$0.83 = i \times 5.12 \times \frac{0.223 \times 1000}{136 \times 4.4}$.
$0.83 = i \times 5.12 \times 0.3728$.
$i = \frac{0.83}{1.908} \approx 0.435$.
Since the van't Hoff factor $i < 1$,the solute undergoes association. For dimerization,$i = 1 - \alpha + \alpha/2 = 1 - \alpha/2$.
$0.435 = 1 - \alpha/2 \implies \alpha/2 = 0.565 \implies \alpha = 1.13$.
Since the value of $i$ is approximately $0.5$,it indicates that phenylacetic acid exists as a dimer in benzene.

(B) The molar mass of $FeSO_4$ is $56 + 32 + 4 \times 16 = 152 \ g/mol$.
Given experimental molecular mass $(M_{exp})$ = $80 \ g/mol$.
The van't Hoff factor $(i)$ is calculated as $i = \frac{M_{theoretical}}{M_{experimental}} = \frac{152}{80} = 1.9$.
For the dissociation of $FeSO_4 \rightleftharpoons Fe^{2+} + SO_4^{2-}$, the number of ions produced $(n)$ is $2$.
The degree of dissociation $(\alpha)$ is given by the formula $\alpha = \frac{i - 1}{n - 1}$.
Substituting the values: $\alpha = \frac{1.9 - 1}{2 - 1} = \frac{0.9}{1} = 0.9$.

(B) Given:
Molality $(m)$ = $0.262 \ mol \ kg^{-1}$
Freezing point of solution $(T_f)$ = $277.4 \ K$
Freezing point of pure benzene $(T_f^\circ)$ = $278.4 \ K$
$K_f$ for benzene = $5.0 \ K \ kg \ mol^{-1}$
Step $1$: Calculate the depression in freezing point $(\Delta T_f)$:
$\Delta T_f = T_f^\circ - T_f = 278.4 \ K - 277.4 \ K = 1.0 \ K$
Step $2$: Calculate the theoretical depression in freezing point $(\Delta T_f(\text{theoretical}))$:
$\Delta T_f(\text{theoretical}) = K_f \times m = 5.0 \ K \ kg \ mol^{-1} \times 0.262 \ mol \ kg^{-1} = 1.31 \ K$
Step $3$: Calculate the van't Hoff factor $(i)$:
$i = \frac{\Delta T_f(\text{observed})}{\Delta T_f(\text{theoretical})} = \frac{1.0}{1.31} \approx 0.763$
Thus,the van't Hoff factor is approximately $0.76$.

(C) Benzoic acid $(C_6H_5COOH)$ undergoes dimerization in benzene due to intermolecular hydrogen bonding.
The reaction is: $2C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$.
Since association occurs,the number of particles decreases,which leads to a decrease in the observed colligative property.
The van't Hoff factor $(i)$ is defined as the ratio of observed colligative property to calculated colligative property.
For association,$i < 1$.

(B) The van't Hoff factor $i$ is given by the ratio of theoretical molar mass to observed molar mass: $i = \frac{M_{\text{theoretical}}}{M_{\text{observed}}} = \frac{164}{65.4} \approx 2.5$.
For the dissociation of $Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$,the number of ions produced per formula unit is $n = 3$.
The relationship between $i$ and the degree of dissociation $\alpha$ is $i = 1 + (n - 1)\alpha$.
Substituting the values: $2.5 = 1 + (3 - 1)\alpha$.
$2.5 = 1 + 2\alpha$.
$1.5 = 2\alpha$.
$\alpha = \frac{1.5}{2} = 0.75$.

(C) For isotonic solutions,the osmotic pressure is equal,so $\pi_1 = \pi_2$.
Since $\pi = iCRT$,and $T$ is constant,we have $i_1 C_1 = i_2 C_2$.
For glucose,$i_2 = 1$ (non-electrolyte),so $i_1 \times 0.02 = 1 \times 0.05$.
$i_1 = \frac{0.05}{0.02} = 2.5$.
For $Pb(NO_3)_2 \rightarrow Pb^{2+} + 2NO_3^-$,the number of ions $n = 3$.
The van't Hoff factor $i = 1 + \alpha(n - 1)$.
$2.5 = 1 + \alpha(3 - 1)$.
$2.5 = 1 + 2\alpha$.
$1.5 = 2\alpha$.
$\alpha = 0.75$.

(C) Acetic acid $(CH_3COOH)$ undergoes dimerization in benzene due to hydrogen bonding: $2CH_3COOH \rightleftharpoons (CH_3COOH)_2$.
The molar mass of acetic acid $(CH_3COOH)$ is $60 \ g/mol$.
For complete association,the van't Hoff factor $(i)$ is $1/2$.
The relation between experimental molecular mass $(M_{exp})$ and theoretical molecular mass $(M_{theo})$ is $M_{exp} = M_{theo} / i$.
Substituting the values: $M_{exp} = 60 / (1/2) = 120 \ g/mol$.

(D) The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$.
For a solute that dimerizes,the van't Hoff factor $i$ is given by $i = 1 - \alpha + \alpha / 2 = 1 - \alpha / 2$,where $\alpha$ is the degree of association.
Since $0 \le \alpha \le 1$,the range for $i$ is $1 - 1/2 \le i \le 1 - 0$,which means $0.5 \le i \le 1$.
Therefore,the range for $\Delta T_b$ is $0.5 mK_b \le \Delta T_b \le 1 mK_b$.
Checking the options:
$A) 0.5 mK_b$ (Possible at $\alpha = 1$)
$B) 0.6 mK_b$ (Possible at $\alpha = 0.8$)
$C) 0.75 mK_b$ (Possible at $\alpha = 0.5$)
$D) 0.33 mK_b$ (Not possible as $i < 0.5$ is required,which is outside the range).

(B) The depression in freezing point is given by the formula: $\Delta T_f = i \times K_f \times m$.
For a trimer,the solute molecules associate to form a single unit,so the van't Hoff factor $i = 1/n$,where $n$ is the number of molecules associating.
Here,$n = 3$,so $i = 1/3$.
Substituting the value of $i$ into the formula: $\Delta T_f = (1/3) \times K_f \times m = mK_f / 3$.

(D) For an $AB$ type ionic substance,the van't Hoff factor $i$ is given by $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation. Since $0 \leq \alpha \leq 1$,the value of $i$ ranges from $1$ to $2$. The ratio of theoretical molecular mass $(M_{th})$ to experimental molecular mass $(M_{exp})$ is equal to the van't Hoff factor $i$ $(i = M_{th} / M_{exp})$. Therefore,the ratio $M_{th} / M_{exp}$ must be in the range $[1, 2]$. Among the given options,$5/2 = 2.5$,which is greater than $2$. Thus,$5 : 2$ is not a consistent ratio.

(D) The boiling point of the solution is $T_b = T_b^o + \Delta T_b = T_b^o + i K_b m$.
The freezing point of the solution is $T_f = T_f^o - \Delta T_f = T_f^o - i K_f m$.
The difference is given as $T_b - T_f = 75.7 \ ^\circ C$.
Substituting the expressions: $(T_b^o + i K_b m) - (T_f^o - i K_f m) = 75.7$.
$(80 + i \times 2.65 \times 0.2) - (5.5 - i \times 5.12 \times 0.2) = 75.7$.
$(80 - 5.5) + i \times 0.2 \times (2.65 + 5.12) = 75.7$.
$74.5 + i \times 0.2 \times (7.77) = 75.7$.
$i \times 1.554 = 75.7 - 74.5 = 1.2$.
$i = \frac{1.2}{1.554} \approx 0.77$.

(A) Colligative properties depend on the number of particles in the solution.
Benzoic acid $(C_6H_5COOH)$ undergoes dimerisation in non-polar solvents like benzene due to hydrogen bonding.
This process reduces the total number of particles in the solution.
Since the colligative property is inversely proportional to the molar mass,a decrease in the number of particles leads to an abnormally high observed molecular mass compared to the theoretical value.

(A) Acetic acid $(CH_3COOH)$ undergoes dimerization in nonpolar solvents like benzene due to hydrogen bonding,resulting in an observed molecular weight that is double the theoretical value.
In polar solvents like water,acetic acid undergoes dissociation into ions ($CH_3COO^-$ and $H^+$),resulting in an observed molecular weight that is lower than the theoretical value.
Since the extent of association or dissociation depends on the polarity of the solvent,the molecular weight determined by the depression in freezing point method is different in benzene and water.
Thus,both the Assertion and Reason are correct,and the Reason correctly explains why the molecular weight differs due to the solvent's polarity.

(D) Given: $w_{2} = 2 \,g$,$K_{f} = 4.9 \,K \,kg \,mol^{-1}$,$w_{1} = 25 \,g$,$\Delta T_{f} = 1.62 \,K$.
The molar mass of benzoic acid $(C_{6}H_{5}COOH)$ is $M_{normal} = (7 \times 12) + (6 \times 1) + (2 \times 16) = 122 \,g \,mol^{-1}$.
Using the formula for depression in freezing point: $\Delta T_{f} = K_{f} \times m = K_{f} \times \frac{w_{2} \times 1000}{M_{observed} \times w_{1}}$.
$M_{observed} = \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} = \frac{4.9 \times 2 \times 1000}{1.62 \times 25} = 241.98 \,g \,mol^{-1}$.
Van't Hoff factor $i = \frac{M_{normal}}{M_{observed}} = \frac{122}{241.98} \approx 0.504$.
For dimerization: $2A \rightleftharpoons A_{2}$.
$i = 1 - x + \frac{x}{2} = 1 - \frac{x}{2}$,where $x$ is the degree of association.
$0.504 = 1 - \frac{x}{2} \implies \frac{x}{2} = 0.496 \implies x = 0.992$.
Percentage association = $0.992 \times 100 = 99.2 \%$.

(N/A) Abnormal molar mass refers to the molar mass of a solute that is either lower or higher than the expected or normal value,calculated based on colligative properties.
$1$. Dissociation: When ionic compounds dissolve in water,they dissociate into cations and anions,increasing the number of particles in the solution. For example,$KCl$ dissociates into $K^{+}$ and $Cl^{-}$ ions. Since colligative properties depend on the number of particles,the observed effect is higher than expected,leading to a lower experimentally determined molar mass compared to the theoretical value.
$2$. Association: Molecules may associate to form larger units,reducing the total number of particles in the solution. $A$ classic example is the dimerization of acetic acid $(CH_3COOH)$ in benzene due to hydrogen bonding,as shown below:
$2 CH_3COOH \rightleftharpoons (CH_3COOH)_2$
In this case,the number of particles is reduced,resulting in a lower observed colligative property value and a higher experimentally determined molar mass compared to the theoretical value.

(C) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
First,calculate the molality $(m)$ of the solution:
Moles of $KCl = \frac{\text{mass}}{\text{molar mass}} = \frac{1.0 \ g}{74.5 \ g \ mol^{-1}} \approx 0.0134 \ mol$.
Mass of solvent in $kg = 200 \ g = 0.2 \ kg$.
Molality $(m) = \frac{0.0134 \ mol}{0.2 \ kg} = 0.067 \ mol \ kg^{-1}$.
Now,substitute the values into the depression in freezing point equation:
$0.24 \ K = i \times 1.86 \ K \ kg \ mol^{-1} \times 0.067 \ mol \ kg^{-1}$.
$i = \frac{0.24}{1.86 \times 0.067} \approx \frac{0.24}{0.1246} \approx 1.926$.
Rounding to the nearest option,the Van't Hoff factor $(i)$ is approximately $2.0$.

(B) $1$. Calculate molality $(m)$: $m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{6.1 \ g}{122 \ g/mol \times 0.5 \ kg} = 0.1 \ m$.
$2$. Calculate depression in freezing point $(\Delta T_f)$: $\Delta T_f = T_f^o - T_f = 5.50 \ ^oC - (-0.290 \ ^oC) = 5.79 \ ^oC$.
$3$. Calculate van't Hoff factor $(i)$: $\Delta T_f = i \times K_f \times m \implies 5.79 = i \times 5.12 \times 0.1 \implies i = \frac{5.79}{0.512} \approx 1.13$.
$4$. For association of benzoic acid: $2C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$.
$5$. The degree of association $(\alpha)$ is given by $i = 1 + \alpha(\frac{1}{n} - 1)$,where $n=2$.
$6$. $1.13 = 1 + \alpha(\frac{1}{2} - 1) \implies 0.13 = -0.5\alpha \implies \alpha = 0.26$.
$7$. Note: The provided solution $95 \%$ is inconsistent with the data provided in the prompt. Based on standard calculation,the result is $26 \%$.

(A) The formula for osmotic pressure is $\pi = iCRT$,where $i$ is the van't Hoff factor.
Given: Mass of $K_2SO_4 = 1.7920 \ g$,Molar mass of $K_2SO_4 = 174.26 \ g/mol$,Volume $V = 1000 \ mL = 1 \ L$,$\pi = 0.680 \ bar$,$T = 26 + 273.15 = 299.15 \ K$,$R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$.
Calculate molarity $C = \frac{n}{V} = \frac{1.7920 \ g}{174.26 \ g/mol \times 1 \ L} \approx 0.01028 \ mol/L$.
Using $\pi = iCRT$,we get $i = \frac{\pi}{CRT} = \frac{0.680}{0.01028 \times 0.08314 \times 299.15} \approx \frac{0.680}{0.2557} \approx 2.66$.

(25 %) $1$. Molar mass of phenol $(C_6H_5OH)$ = $94 \ g \ mol^{-1}$.
$2$. Molality $(m)$ = $\frac{1.5 \ g / 94 \ g \ mol^{-1}}{0.1 \ kg} = 0.1596 \ mol \ kg^{-1}$.
$3$. Observed depression in freezing point: $\Delta T_f = K_f \times m \times i$.
$4$. $0.56 = 4 \times 0.1596 \times i \implies i = \frac{0.56}{0.6384} \approx 0.877$.
$5$. For dimerization,$i = 1 + (\frac{1}{n} - 1)\alpha$,where $n = 2$.
$6$. $0.877 = 1 + (\frac{1}{2} - 1)\alpha \implies 0.877 = 1 - 0.5\alpha$.
$7$. $0.5\alpha = 0.123 \implies \alpha = 0.246 \approx 25 \%$.

(N/A) Association: It is the process in which two or more molecules of a solute combine to form a larger single molecule. This leads to a decrease in the total number of particles in the solution,resulting in a van't Hoff factor $(i < 1)$. For example,the dimerization of acetic acid in benzene.
Dissociation: It is the process in which a solute molecule splits into two or more simpler ions or smaller molecules when dissolved in a solvent. This leads to an increase in the total number of particles in the solution,resulting in a van't Hoff factor $(i > 1)$. For example,the dissociation of $NaCl$ into $Na^+$ and $Cl^-$ ions in water.

(N/A) Abnormal molar mass is defined as the molar mass of a substance that is either higher or lower than the expected (theoretical) value calculated based on its chemical formula.
This phenomenon occurs due to the association or dissociation of solute particles in a solution.
$1$. Association: When solute particles combine to form larger aggregates,the number of particles decreases,leading to a higher observed molar mass than the theoretical value.
$2$. Dissociation: When solute particles break apart into ions,the number of particles increases,leading to a lower observed molar mass than the theoretical value.
The extent of this deviation is expressed using the van't Hoff factor $(i)$.

(C) The dimerisation reaction is: $2 C_{6}H_{5}COOH \rightleftharpoons (C_{6}H_{5}COOH)_{2}$
Initially,we have $1$ mole of benzoic acid.
After association,the moles are:
Benzoic acid: $(1-x)$
Dimer: $x/2$
Total moles after association $= (1-x) + (x/2) = 1 - x/2$
The van't Hoff factor $(i)$ is defined as the ratio of total moles after association to the initial moles:
$i = \frac{1 - x/2}{1} = 1 - \frac{x}{2}$

(A) The freezing point depression is given by $\Delta T_f = T_f^{\circ} - T_f = 0^{\circ} C - (-3.885^{\circ} C) = 3.885 \ K$.
For a weak acid $HA$ undergoing dissociation,$HA \rightleftharpoons H^+ + A^-$,the van't Hoff factor $i = 1 + \alpha$.
Using the formula $\Delta T_f = i \times K_f \times m$:
$3.885 = (1 + \alpha) \times 1.85 \times 2$.
$3.885 = (1 + \alpha) \times 3.7$.
$1 + \alpha = \frac{3.885}{3.7} = 1.05$.
$\alpha = 1.05 - 1 = 0.05$.
Expressing $\alpha$ in terms of $10^{-3}$:
$\alpha = 0.05 = 50 \times 10^{-3}$.
Thus,the value is $50$.

(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0} = 100.52^{\circ} C - 100^{\circ} C = 0.52^{\circ} C$.
For dimerization,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
Using the formula $\Delta T_{b} = i \times K_{b} \times m$,where $m = 2 \, M$:
$0.52 = (1 - \frac{\alpha}{2}) \times 0.52 \times 2$.
Dividing both sides by $0.52$,we get $1 = (1 - \frac{\alpha}{2}) \times 2$.
$1 = 2 - \alpha$.
$\alpha = 1$.
Thus,the percentage association is $1 \times 100 = 100 \, \%$.

(C) The freezing point depression is given by $\Delta T_{f} = i \times K_{f} \times m$.
Given $\Delta T_{f} = 0 - (-0.93) = 0.93 \ K$.
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}} = \frac{12.2}{122} \times \frac{1000}{100} = 1 \ m$.
Substituting values: $0.93 = i \times 1.86 \times 1$,which gives $i = 0.5$.
For association,the van't Hoff factor is $i = 1 + (\frac{1}{n} - 1)\alpha$.
Since $\alpha = 1$ ($100 \%$ association),$0.5 = 1 + (\frac{1}{n} - 1) \times 1$.
$0.5 = \frac{1}{n}$,therefore $n = 2$.

(A) Let the initial moles of $HA$ be $a = 1 \text{ mole}$.
$50\,\%$ of $HA$ $(0.5 \text{ mole})$ dimerizes: $2HA \rightleftharpoons H_2A_2$.
Final moles for dimerization: $0.5 - 0.25 = 0.25 \text{ mole of } HA$ and $0.25 \text{ mole of } H_2A_2$.
$50\,\%$ of $HA$ $(0.5 \text{ mole})$ dissociates: $HA \rightleftharpoons H^+ + A^-$.
Final moles for dissociation: $0.5 \text{ mole of } H^+$ and $0.5 \text{ mole of } A^-$.
Total final moles $= 0.25 (HA) + 0.25 (H_2A_2) + 0.5 (H^+) + 0.5 (A^-) = 1.5 \text{ moles}$.
Van't Hoff factor $i = \frac{\text{Total final moles}}{\text{Initial moles}} = \frac{1.5}{1} = 1.5$.
$1.5 = 150 \times 10^{-2}$.

(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$.
First,calculate the molality $(m)$: $m = \frac{0.7 \ g}{93 \ g \ mol^{-1}} \times \frac{1000 \ g \ kg^{-1}}{42.0 \ g} = 0.1792 \ mol \ kg^{-1}$.
Using $\Delta T_f = 0.2 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$,we find the van't Hoff factor $(i)$:
$0.2 = i \cdot 1.86 \cdot 0.1792 \implies i = \frac{0.2}{1.86 \cdot 0.1792} \approx 0.60$.
For the association reaction $2A \rightleftharpoons A_2$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
Equating $1 - \frac{\alpha}{2} = 0.60$,we get $\frac{\alpha}{2} = 0.40$,so $\alpha = 0.80$.
The percentage association is $80 \%$.

(D) Given: $0.5\%$ solution of $KCl$ means $0.5 \, g$ of $KCl$ in $99.5 \, g$ of water.
Molality $(m) = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in } g} = \frac{0.5}{74.6} \times \frac{1000}{99.5} \approx 0.0673 \, mol \, kg^{-1}$.
Using the formula for depression in freezing point: $\Delta T_f = i \times K_f \times m$.
$0.24 = i \times 1.80 \times 0.0673$.
$i = \frac{0.24}{1.80 \times 0.0673} \approx 1.981$.
For $KCl \rightarrow K^+ + Cl^-$,the number of ions $(n) = 2$.
Degree of dissociation $(\alpha) = \frac{i - 1}{n - 1} = \frac{1.981 - 1}{2 - 1} = 0.981$.
Percentage dissociation $= 0.981 \times 100 = 98.1\%$.
The nearest integer is $98\%$ (or $99\%$ depending on rounding precision,here $98$ is closer).

(B) $1$. Calculate the mass of acetic acid: $Mass = Density \times Volume = 1.02 \, g \, mL^{-1} \times 1.2 \, mL = 1.224 \, g$.
$2$. Calculate the moles of acetic acid: $Moles = \frac{1.224 \, g}{60 \, g \, mol^{-1}} = 0.0204 \, mol$.
$3$. Calculate molality $(m)$: Since the solution is dilute,we assume the density of the solution is approximately $1 \, g \, mL^{-1}$,so $2 \, L$ of solution is $2 \, kg$ of water. $m = \frac{0.0204 \, mol}{2 \, kg} = 0.0102 \, mol \, kg^{-1}$.
$4$. Use the depression in freezing point formula: $\Delta T_{f} = i \times K_{f} \times m$.
$5$. For acetic acid dissociation $(CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+})$,$i = 1 + \alpha$.
$6$. Substitute values: $0.0198 = (1 + \alpha) \times 1.85 \times 0.0102$.
$7$. Solve for $\alpha$: $1 + \alpha = \frac{0.0198}{1.85 \times 0.0102} \approx 1.04928$.
$8$. $\alpha = 0.04928$,which is $4.928 \% \approx 5 \%$. The nearest integer is $5$.

(C) Concentration of $HCOOH = 0.5 \, mL \, L^{-1}$.
Mass of $HCOOH = 0.5 \, mL \times 1.05 \, g \, mL^{-1} = 0.525 \, g \, L^{-1}$.
Molar mass of $HCOOH = 46 \, g \, mol^{-1}$.
Molality $(m) = \frac{0.525 \, g}{46 \, g \, mol^{-1} \times 1 \, kg} \approx 0.01141 \, mol \, kg^{-1}$.
Using the formula $\Delta T_{f} = i K_{f} m$,we have $i = \frac{\Delta T_{f}}{K_{f} m}$.
$i = \frac{0.0405}{1.86 \times 0.01141} \approx \frac{0.0405}{0.02122} \approx 1.908$.
Thus,the Van't Hoff factor is nearly $1.9$.

(B) For a weak monobasic acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$.
The van't Hoff factor $i = 1 + \alpha$,where $\alpha = 0.3$.
So,$i = 1 + 0.3 = 1.3$.
The observed freezing point depression is $(\Delta T_f)_{obs} = i \times K_f \times m$.
The theoretical freezing point depression is $(\Delta T_f)_{cal} = K_f \times m$.
The percentage increase in freezing point depression is given by $\frac{(\Delta T_f)_{obs} - (\Delta T_f)_{cal}}{(\Delta T_f)_{cal}} \times 100$.
Substituting the values: $\frac{i \times K_f \times m - K_f \times m}{K_f \times m} \times 100 = (i - 1) \times 100$.
Percentage increase $= (1.3 - 1) \times 100 = 0.3 \times 100 = 30\%$.

(B) The molar mass of phenol $(C_6H_5OH)$ is $M = 94 \ g \ mol^{-1}$.
Given mass of phenol $w = 75.2 \ g$,so moles $n = \frac{75.2}{94} = 0.8 \ mol$.
Assuming $1 \ kg$ of solvent,molality $m = 0.8 \ mol \ kg^{-1}$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Given $\Delta T_f = 7 \ K$ and $K_f = 14 \ K \ kg \ mol^{-1}$,we have $7 = i \times 14 \times 0.8$.
$i = \frac{7}{14 \times 0.8} = \frac{1}{1.6} = 0.625$.
For dimerization,$i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
$0.625 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.375 \implies \alpha = 0.75$.
Thus,the percentage of phenol that dimerises is $75\%$.

(A) The formula for freezing point depression is $\Delta T_{f} = K_{f} \times m \times i$,where $m$ is the molality.
First,calculate the molar mass of naphthoic acid $(C_{11}H_8O_2)$: $(11 \times 12) + (8 \times 1) + (2 \times 16) = 132 + 8 + 32 = 172 \ g \ mol^{-1}$.
Calculate the molality $(m)$ of the solution: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{20 \ g / 172 \ g \ mol^{-1}}{0.050 \ kg} = \frac{20}{172 \times 0.050} = \frac{20}{8.6} \approx 2.325 \ mol \ kg^{-1}$.
Substitute the values into the equation: $2 = 1.72 \times \left( \frac{20}{172 \times 0.050} \right) \times i$.
$2 = 1.72 \times \left( \frac{20}{8.6} \right) \times i$.
$2 = 1.72 \times 2.3255 \times i$.
$2 = 4.0 \times i$.
$i = \frac{2}{4} = 0.5$.
Therefore,the van't Hoff factor $(i)$ is $0.5$,and option $(A)$ is correct.

(A) The dissociation of $MX_2$ is given by: $MX_2 \rightarrow M^{2+} + 2X^-$.
The van't Hoff factor $i$ is calculated as: $i = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} = \frac{164}{65.6} = 2.5$.
For the dissociation $MX_2 \rightarrow M^{2+} + 2X^-$,the number of ions produced per formula unit is $n = 3$.
The relation between $i$ and degree of dissociation $\alpha$ is: $i = 1 + (n - 1)\alpha$.
Substituting the values: $2.5 = 1 + (3 - 1)\alpha$.
$2.5 = 1 + 2\alpha$.
$1.5 = 2\alpha$.
$\alpha = 0.75$.
Therefore,the percent degree of ionisation is $0.75 \times 100 = 75 \%$.

(A) The formula for freezing point depression is $\Delta T_{f} = i K_{f} m$.
Given $\Delta T_{f} = 0.20^{\circ} C$,$K_{f} = 1.8 \ K \ kg \ mol^{-1}$,and $m = 0.1 \ m$.
$0.20 = i \times 1.8 \times 0.1$
$i = \frac{0.20}{0.18} = \frac{10}{9}$.
For the dissociation $HA \rightleftharpoons H^{+} + A^{-}$,the van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
$1 + \alpha = \frac{10}{9} \implies \alpha = \frac{1}{9}$.
The dissociation constant $K_{a} = \frac{C \alpha^2}{1 - \alpha}$.
Substituting $C = 0.1$ and $\alpha = \frac{1}{9}$:
$K_{a} = \frac{0.1 \times (1/9)^2}{1 - 1/9} = \frac{0.1 \times (1/81)}{8/9} = \frac{0.1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 \times 10^{-3}$.

(D) For the dissociation of $K_4[Fe(CN)_6]$:
$K_4[Fe(CN)_6] \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$
Here,the number of ions produced per formula unit is $n = 5$.
Using the formula for the van't Hoff factor $(i)$ in terms of degree of dissociation $(x)$:
$i = 1 + (n - 1)x = 1 + (5 - 1)x = 1 + 4x$.
We know that the van't Hoff factor is also defined as the ratio of normal molecular weight to abnormal molecular weight:
$i = \frac{m_{normal}}{m_{abnormal}}$.
Therefore,$m_{abnormal} = \frac{m_{normal}}{i} = \frac{m}{1 + 4x}$.

(B) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
Given: $\Delta T_f = 0.046 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$m = 0.02 \ m$.
First,calculate the van't Hoff factor $(i)$:
$i = \frac{\Delta T_f}{K_f \times m} = \frac{0.046}{1.86 \times 0.02} = \frac{0.046}{0.0372} \approx 1.2366$.
For dissociation,the degree of dissociation $(\alpha)$ is given by $\alpha = \frac{i - 1}{n - 1}$.
Given $n = 2$,$\alpha = \frac{1.2366 - 1}{2 - 1} = 0.2366$.
Percent dissociation = $\alpha \times 100 = 0.2366 \times 100 = 23.66 \% \approx 23.6 \%$.
Therefore,the correct option is $B$.

(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.
Given: $\Delta T_{f} = 0 - (-0.54) = 0.54 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.18 \ m$.
Substituting the values: $0.54 = i \times 1.86 \times 0.18$.
Calculating $i$: $i = \frac{0.54}{1.86 \times 0.18} = \frac{0.54}{0.3348} \approx 1.612$.

(D) The formula for freezing point depression is $\Delta T_{f} = i \times m \times K_{f}$.
Given $\Delta T_{f} = 0.7 \ K$,$m = 0.2 \ m$,and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.7 = i \times 0.2 \times 1.86$.
$i = \frac{0.7}{0.2 \times 1.86} = \frac{0.7}{0.372} \approx 1.882$.

(D) The depression in freezing point is given by the formula: $\Delta T_{f} = i \times m \times K_{f}$
Given: $\Delta T_{f} = 0 - (-0.680^{\circ} C) = 0.680 \ K$,molality $m = 0.2 \ mol \ kg^{-1}$,and $K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.680 = i \times 0.2 \times 1.86$
$i = \frac{0.680}{0.2 \times 1.86} = \frac{0.680}{0.372} \approx 1.8279$
Rounding to two decimal places,we get $i \approx 1.83$.

(D) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{0} - T_{f} = 0 - (-0.5 \ K) = 0.5 \ K$.
Using the formula for depression in freezing point with the van't Hoff factor: $\Delta T_{f} = i \times K_{f} \times m$.
Assuming molarity $(M)$ is approximately equal to molality $(m)$ for dilute aqueous solutions,$m \approx 0.15 \ mol \ kg^{-1}$.
Rearranging for $i$: $i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.5 \ K}{1.86 \ K \ kg \ mol^{-1} \times 0.15 \ mol \ kg^{-1}}$.
Calculating the value: $i = \frac{0.5}{0.279} \approx 1.79$.

(C) The depression in freezing point is given by $\Delta T_{f} = T_{f}^0 - T_{f}$.
Given $T_{f}^0 = 0^{\circ} C$ and $T_{f} = -0.51^{\circ} C$,so $\Delta T_{f} = 0 - (-0.51) = 0.51 \ K$.
The formula for depression in freezing point is $\Delta T_{f} = i \times K_{f} \times m$.
Substituting the values: $0.51 = i \times 1.86 \times 0.15$.
Solving for $i$: $i = \frac{0.51}{1.86 \times 0.15} = \frac{0.51}{0.279} \approx 1.828$.
Rounding to two decimal places,the van't Hoff factor $i$ is $1.82$.

(C) The formula for depression in freezing point is $\Delta T_{f} = i \times K_{f} \times m$.
Given: $\Delta T_{f} = 0 - (-0.056) = 0.056 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$.
Substituting the values: $0.056 = i \times 1.86 \times 0.01$.
Therefore,$i = \frac{0.056}{1.86 \times 0.01} = \frac{0.056}{0.0186} \approx 3.01$.

(B) Given: Molality $(m) = 0.2 \ m$,Freezing point of solution $(T_f) = -0.680 \ ^{\circ}C$,Freezing point of pure water $(T_f^{\circ}) = 0 \ ^{\circ}C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Depression in freezing point $(\Delta T_f) = T_f^{\circ} - T_f = 0 - (-0.680) = 0.680 \ K$.
The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Substituting the values: $0.680 = i \times 1.86 \times 0.2$.
$i = \frac{0.680}{1.86 \times 0.2} = \frac{0.680}{0.372} \approx 1.8279$.
Rounding to two decimal places,the Van't Hoff factor $(i) \approx 1.83$.

(B) The Van't Hoff factor '$i$' is defined as the ratio of the observed colligative property to the calculated colligative property.
It accounts for the extent of dissociation or association of solute particles in a solution.

(A) Dimerisation of solute molecules in low dielectric constant solvents is primarily due to the formation of intermolecular $hydrogen \ bonds$.
In solvents with low dielectric constants,the electrostatic screening is weak,which facilitates the association of solute molecules through $hydrogen \ bonding$ to form stable dimers,such as the dimerization of carboxylic acids.

(D) Given: Mass of solute $w_{B} = 1.25 \ g$,Mass of solvent $w_{A} = 50 \ g$,$\Delta T_{f} = 0.3 \ K$,$M_{\text{normal}} = 94 \ g \ mol^{-1}$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.
First,calculate the observed molar mass $(M_{\text{obs}})$ using the formula: $\Delta T_{f} = K_{f} \times m = K_{f} \times \frac{w_{B} \times 1000}{M_{\text{obs}} \times w_{A}}$.
$M_{\text{obs}} = \frac{1.86 \times 1.25 \times 1000}{0.3 \times 50} = 155 \ g \ mol^{-1}$.
Calculate the van't Hoff factor $(i)$: $i = \frac{M_{\text{normal}}}{M_{\text{obs}}} = \frac{94}{155} \approx 0.6064$.
For association,$n P \rightleftharpoons P_{n}$. Assuming dimerization $(n=2)$,the degree of association $\alpha$ is given by $i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha(1 - \frac{1}{n})$.
$0.6064 = 1 - \alpha(1 - 0.5) \implies 0.6064 = 1 - 0.5\alpha$.
$0.5\alpha = 1 - 0.6064 = 0.3936$.
$\alpha = 0.7872$ or approximately $78.7\%$.
Since $78.7\%$ is not among the options,the correct choice is $D$.