Abnormal molecular mass Questions in English

(A) The dissociation reaction for $BaCl_2$ is: $BaCl_2 \rightleftharpoons Ba^{2+} + 2Cl^-$
Let the initial moles be $1$. After dissociation,the moles are: $BaCl_2 = 1 - \alpha$,$Ba^{2+} = \alpha$,$Cl^- = 2\alpha$,where $\alpha$ is the degree of dissociation.
The total number of particles after dissociation is $n = (1 - \alpha) + \alpha + 2\alpha = 1 + 2\alpha$.
The Van't Hoff factor $i$ is given by the ratio of total particles to initial particles: $i = 1 + 2\alpha$.
Given $i = 1.98$,we have: $1.98 = 1 + 2\alpha$.
$2\alpha = 1.98 - 1 = 0.98$.
$\alpha = 0.49$.
Percentage dissociation = $\alpha \times 100 = 0.49 \times 100 = 49\%$.

(C) For isotonic solutions,the osmotic pressure is equal,so the van't Hoff factor multiplied by concentration is equal: $i_1 C_1 = i_2 C_2$.
For glucose (non-electrolyte),$i_2 = 1$,so $i_2 C_2 = 1 \times 0.010 = 0.010 \ M$.
For $Na_2SO_4$,the dissociation is $Na_2SO_4 \rightleftharpoons 2Na^{+} + SO_4^{2-}$.
Let $\alpha$ be the degree of dissociation. The van't Hoff factor $i_1 = 1 + (n-1)\alpha$,where $n=3$.
So,$i_1 = 1 + (3-1)\alpha = 1 + 2\alpha$.
The condition $i_1 C_1 = 0.010$ becomes $(1 + 2\alpha) \times 0.004 = 0.010$.
$1 + 2\alpha = \frac{0.010}{0.004} = 2.5$.
$2\alpha = 1.5$.
$\alpha = 0.75$.
Therefore,the percentage of dissociation is $0.75 \times 100 = 75 \%$.

(C) We know that $NaCl$ is a strong electrolyte that dissociates completely in water as $NaCl \rightarrow Na^+ + Cl^-$.
The van't Hoff factor $(i)$ for $NaCl$ is $2$.
The relationship between the van't Hoff factor and molar mass is given by $i = \frac{\text{theoretical molar mass}}{\text{observed molar mass}}$.
Since $i = 2$,the observed (experimental) molar mass is $\frac{1}{2}$ of the theoretical molar mass.
Therefore,the molecular weight determined by the osmotic pressure method will be lower than the theoretical value.

(A) Benzoic acid $(C_6H_5COOH)$ in benzene undergoes dimerization due to intermolecular hydrogen bonding.
The molecular mass of benzoic acid is $122 \ g/mol$.
Due to dimerization,the observed molecular mass becomes double the normal molecular mass,which is $2 \times 122 = 244 \ g/mol$.

(A) In the case of association of solute particles in a solution,the observed molecular weight is higher than the normal molecular weight,resulting in a Van't Hoff factor $i < 1$.
In the case of dissociation of solute particles,the observed molecular weight is lower than the normal molecular weight,resulting in a Van't Hoff factor $i > 1$.
Therefore,the Van't Hoff factor calculated from association data is always less than that calculated from dissociation data.

(C) The dissociation of $Na_2SO_4$ is represented as: $Na_2SO_4 \rightleftharpoons 2Na^{+} + SO_4^{2-}$
Initial moles: $1 \quad 0 \quad 0$
Moles at equilibrium: $1 - \alpha \quad 2\alpha \quad \alpha$
Total moles at equilibrium = $(1 - \alpha) + 2\alpha + \alpha = 1 + 2\alpha$
The Van't Hoff factor $(i)$ is defined as the ratio of the total number of particles after dissociation to the initial number of particles.
$i = \frac{1 + 2\alpha}{1} = 1 + 2\alpha$

(A) The Van't Hoff factor $(i)$ is defined as the ratio of the normal (theoretical) molecular mass to the observed (experimental) molecular mass of a solute.
$i = \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}}$
For solutes that undergo association,the number of particles decreases,so $i < 1$.
For solutes that undergo dissociation,the number of particles increases,so $i > 1$.

(D) Urea is a non-electrolyte solute.
It does not undergo dissociation or association in an aqueous solution.
Therefore,the Van't Hoff factor $i$ for urea is $1.0$.

(C) For the association reaction $nA \rightleftharpoons (A)_n$,let the degree of association be $x$.
Initial moles: $A = 1$,$(A)_n = 0$.
Moles at equilibrium: $A = 1 - x$,$(A)_n = \frac{x}{n}$.
Total moles at equilibrium = $(1 - x) + \frac{x}{n}$.
The Van't Hoff factor $i$ is defined as the ratio of total moles of particles after association to the initial moles of solute.
$i = \frac{1 - x + \frac{x}{n}}{1} = 1 - x + \frac{x}{n}$.

(B) The molecular weight of acetic acid $(CH_3COOH)$ is $60 \ g/mol$.
In a non-polar solvent like benzene,acetic acid undergoes intermolecular hydrogen bonding to form a dimer,$(CH_3COOH)_2$.
Due to this association,the observed molecular weight becomes double the normal molecular weight.
Therefore,the observed molecular weight $= 2 \times 60 = 120 \ g/mol$.

(B) In benzene,benzoic acid molecules undergo association to form dimers due to intermolecular hydrogen bonding.
This association reduces the total number of particles in the solution.
Since osmotic pressure is a colligative property,it depends on the number of solute particles.
As the number of particles decreases due to association,the observed osmotic pressure becomes less than the expected value.
The structure of the dimer is as follows:
(Image shows two benzoic acid molecules linked by two hydrogen bonds between the carboxyl groups).

(A) In water,$CH_3COOH$ acts as a weak electrolyte and undergoes partial dissociation into $CH_3COO^-$ and $H^+$ ions,resulting in a lower observed molecular mass $(60)$.
In benzene (a non-polar solvent),$CH_3COOH$ molecules form intermolecular hydrogen bonds,leading to the formation of dimers. This association results in an observed molecular mass of $120$ (double the normal mass).

(B) In benzene,benzoic acid molecules undergo intermolecular hydrogen bonding to form dimers.
This association leads to a decrease in the number of particles in the solution.
Consequently,the observed molecular weight is higher than the theoretical value,which corresponds to the dimerization of benzoic acid.

(C) For isotonic solutions, the osmotic pressure $(\pi)$ is equal, so the molar concentration $(C)$ must be equal: $C_{NaCl} = C_{glucose}$.
Assuming the mass of the solvent is the same for both solutions, the number of moles must be equal.
Moles of glucose = $\frac{7.2 \ g}{180 \ g/mol} = 0.04 \ mol$.
Moles of $NaCl$ = $\frac{1.2 \ g}{58.5 \ g/mol} \approx 0.0205 \ mol$.
Since $NaCl$ dissociates as $NaCl \rightarrow Na^{+} + Cl^{-}$, the van’t Hoff factor $(i)$ is involved in the effective concentration.
$\pi_{NaCl} = i \times C_{NaCl} \times R \times T$ and $\pi_{glucose} = C_{glucose} \times R \times T$.
Equating them: $i \times (0.0205) = 0.04$.
$i = \frac{0.04}{0.0205} \approx 1.95$.

(C) Acetic acid $(CH_3COOH)$ undergoes intermolecular hydrogen bonding in non-polar solvents like benzene to form a dimer.
Due to the formation of the dimer $(CH_3COOH)_2$,the observed molecular mass becomes double the theoretical molecular mass.
The theoretical molecular mass of acetic acid is $60 \ g/mol$,so the observed molecular mass is $2 \times 60 = 120 \ g/mol$.

(B) When $CH_3COOH$ is dissolved in a non-polar solvent like benzene,it undergoes intermolecular hydrogen bonding to form a dimer.
This association of two molecules results in an increase in the observed molecular mass compared to the theoretical molecular mass.
The reaction is: $2CH_3COOH \rightleftharpoons (CH_3COOH)_2$.

(D) In water,benzoic acid $(C_6H_5COOH)$ undergoes dissociation to form benzoate ions and hydronium ions,which leads to a decrease in the observed molecular weight.
In benzene (a non-polar solvent),benzoic acid molecules form intermolecular hydrogen bonds,leading to the formation of a dimer. This association results in an increase in the observed molecular weight.
Therefore,the correct option is $(D)$.

(C) The van't Hoff factor $i$ is given by the ratio of the calculated molecular weight to the observed molecular weight.
$i = \frac{\text{Calculated molecular weight}}{\text{Observed molecular weight}} = \frac{164}{65.6} = 2.5$
For the dissociation of calcium nitrate: $Ca(NO_3)_2 \rightarrow Ca^{2+} + 2NO_3^-$,the number of ions produced per formula unit is $n = 3$.
The relationship between the van't Hoff factor and the degree of dissociation $\alpha$ is $i = 1 + (n - 1)\alpha$.
Substituting the values: $2.5 = 1 + (3 - 1)\alpha$
$2.5 = 1 + 2\alpha$
$1.5 = 2\alpha$
$\alpha = 0.75$
Therefore,the degree of dissociation in percentage is $\alpha \times 100 = 75\%$.

(D) Benzoic acid undergoes association in benzene to form a dimer due to hydrogen bonding.
The molecular weight of benzoic acid is $122 \ g/mol$.
Since it forms a dimer,the observed molecular weight is $2 \times 122 = 244 \ g/mol$.

(B) Molality $(m) = \frac{75.2}{94} = 0.8 \ mol \ kg^{-1}$.
Using the formula for depression in freezing point: $\Delta T_f = i \times K_f \times m$.
$7 = i \times 14 \times 0.8$.
$i = \frac{7}{14 \times 0.8} = \frac{7}{11.2} = 0.625$.
For association,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{n}$.
For dimerization,$n = 2$,so $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
$0.625 = 1 - \frac{\alpha}{2}$.
$\frac{\alpha}{2} = 1 - 0.625 = 0.375$.
$\alpha = 0.375 \times 2 = 0.75$.
Percentage of association = $\alpha \times 100 = 75\%$.

(C) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
First,calculate the molality $(m)$:
$m = \frac{w_2 \times 1000}{M_2 \times w_1} = \frac{0.85 \times 1000}{136.3 \times 125} = 0.0499 \ mol \ kg^{-1}$.
Given $\Delta T_f = 0.23 \ K$,so $0.23 = i \times 1.86 \times 0.0499$.
$i = \frac{0.23}{1.86 \times 0.0499} \approx 2.478$.
For $ZnCl_2 \rightarrow Zn^{2+} + 2Cl^-$,the number of ions $n = 3$.
The degree of dissociation $\alpha = \frac{i - 1}{n - 1} = \frac{2.478 - 1}{3 - 1} = \frac{1.478}{2} = 0.739$.
Thus,$\alpha \approx 73.9\%$,which is closest to $73.5\%$.

(D) $\Delta T_f = T_o - T_f = 5.3 - 4.47 = 0.83 \ K$
Molar mass of $C_6H_5CH_2COOH = 136 \ g/mol$
Molality $(m) = \frac{0.223 \ g}{136 \ g/mol} \times \frac{1000}{4.49 \ g} = 0.365 \ mol/kg$
Using $\Delta T_f = i \times K_f \times m$:
$i = \frac{\Delta T_f}{K_f \times m} = \frac{0.83}{5.12 \times 0.365} \approx 0.444$
Since $i \approx 0.5$,it indicates that the solute molecules associate to form a dimer in benzene.

(B) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$.
Here,the number of ions produced per formula unit,$n = 3$.
The formula for the van't Hoff factor $(i)$ is: $i = 1 + (n - 1)\alpha$,where $\alpha$ is the degree of dissociation.
Substituting the given values: $2.74 = 1 + (3 - 1)\alpha$.
$2.74 = 1 + 2\alpha$.
$2\alpha = 1.74$.
$\alpha = 0.87$.
Therefore,the degree of dissociation in percentage is $0.87 \times 100 = 87\%$.

(D) The association reaction is given by: $3A \rightleftharpoons A_3$.
Assuming $100\%$ association,the number of particles decreases.
The van't Hoff factor $i$ is defined as the ratio of the observed colligative property to the calculated colligative property.
For association,$i = 1 + (1/n - 1)\alpha$,where $n = 3$ and $\alpha = 1$.
Thus,$i = 1 + (1/3 - 1)(1) = 1/3$.

(B) The degree of dissociation $(\alpha)$ is related to the van't Hoff factor $(i)$ by the formula: $\alpha = \frac{i - 1}{n - 1}$.
For $AgNO_3$,the dissociation reaction is $AgNO_3 \rightarrow Ag^+ + NO_3^-$,so $n = 2$.
The van't Hoff factor $(i)$ is the ratio of the calculated molecular mass to the observed molecular mass: $i = \frac{\text{Calculated mass}}{\text{Observed mass}} = \frac{170}{92.64} \approx 1.835$.
Substituting the values into the formula: $\alpha = \frac{1.835 - 1}{2 - 1} = 0.835$.
To express this as a percentage: $\alpha \times 100 = 0.835 \times 100 = 83.5 \%$.

(A) For the dissociation of a weak electrolyte $A_xB_y$:
$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
Initial moles: $1, 0, 0$
Moles at equilibrium: $(1 - \alpha), x\alpha, y\alpha$
Total moles at equilibrium = $(1 - \alpha) + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$
Van't Hoff factor $i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = 1 + \alpha(x + y - 1)$
$i - 1 = \alpha(x + y - 1)$
$\alpha = \frac{i - 1}{x + y - 1}$

(B) In non-polar solvents like benzene,benzoic acid molecules undergo intermolecular hydrogen bonding to form a dimer.
This association leads to a decrease in the number of particles in the solution,which affects the colligative properties used to determine the molar mass.

(C) The van't Hoff factor $i$ is defined as the ratio of the observed colligative property to the calculated colligative property.
For a compound that undergoes dissociation,the number of particles in the solution increases,so $i > 1$.
For a compound that undergoes association,the number of particles in the solution decreases,so $i < 1$.
Therefore,for dissociation and association respectively,$i$ is greater than $1$ and less than $1$.

(A) For the dissociation of a weak electrolyte $A_xB_y$:
$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
At $t=0$,the moles are $1, 0, 0$ respectively.
At equilibrium,the moles are $(1 - \alpha), x\alpha, y\alpha$.
The total number of moles at equilibrium is:
$n_{total} = 1 - \alpha + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$
The van't Hoff factor $(i)$ is defined as the ratio of observed moles to initial moles:
$i = \frac{1 + \alpha(x + y - 1)}{1} = 1 + \alpha(x + y - 1)$
Rearranging for $\alpha$:
$i - 1 = \alpha(x + y - 1)$
$\alpha = \frac{i - 1}{x + y - 1}$

(D) The association reaction is: $2 \ CH_{3}COOH \rightleftharpoons (CH_{3}COOH)_{2}$
Initial moles: $1 \quad 0$
At equilibrium: $1 - \alpha \quad \alpha / 2$
Total moles at equilibrium: $1 - \alpha + \alpha / 2 = 1 - \alpha / 2$
Van't Hoff factor $i = 1 - \alpha / 2$
Using the formula: $\Delta T_{f} = i \times K_{f} \times m$
Molality $m = \frac{0.2 / 60}{20 / 1000} = \frac{0.2 \times 1000}{60 \times 20} = \frac{200}{1200} = 0.1667 \ m$
$0.45 = (1 - \alpha / 2) \times 5.12 \times 0.1667$
$0.45 = (1 - \alpha / 2) \times 0.8535$
$1 - \alpha / 2 = 0.5272$
$\alpha / 2 = 0.4728$
$\alpha = 0.9456$
Percentage association $= 94.56 \% \approx 94.6 \%$

(B) Given: $W_B = 2 \ g$,$K_f = 4.9 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 1.62 \ K$.
First,calculate the observed molar mass $(M_{obs})$ using the formula: $\Delta T_f = K_f \times \frac{W_B}{M_{obs}} \times \frac{1000}{W_A}$.
$M_{obs} = \frac{4.9 \times 2 \times 1000}{1.62 \times 25} = 241.98 \ g \ mol^{-1}$.
The theoretical molar mass of benzoic acid $(M_{theo})$ is $122 \ g \ mol^{-1}$.
The van't Hoff factor $i = \frac{M_{theo}}{M_{obs}} = \frac{122}{241.98} \approx 0.504$.
For the association reaction $2 C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$,the degree of association $\alpha$ is related to $i$ by $i = 1 - \alpha + \frac{\alpha}{n}$,where $n = 2$.
$0.504 = 1 - \alpha + \frac{\alpha}{2} \Rightarrow 0.504 = 1 - \frac{\alpha}{2}$.
$\frac{\alpha}{2} = 1 - 0.504 = 0.496 \Rightarrow \alpha = 0.992$.
Thus,the percentage of benzoic acid in dimeric form is $99.2 \ \%$.

(A) Given: Mass of solute $(w_2)$ = $2\,g$,Mass of solvent $(w_1)$ = $25\,g$,$\Delta T_f = 1.62\,K$,$K_f = 4.9\,K\,kg\,mol^{-1}$,Molar mass of benzoic acid $(M_2)$ = $122\,g\,mol^{-1}$.
We know,$\Delta T_f = i \times K_f \times m$.
$1.62 = i \times 4.9 \times \frac{2 \times 1000}{122 \times 25}$.
$1.62 = i \times 3.213$.
$i = \frac{1.62}{3.213} = 0.5042$.
Benzoic acid exists as a dimer $(n=2)$ in benzene: $i = 1 - \alpha + \frac{\alpha}{n} = 1 - \frac{\alpha}{2}$.
$0.5042 = 1 - \frac{\alpha}{2}$.
$\frac{\alpha}{2} = 0.4958$.
$\alpha = 0.9916$.
Percentage degree of association = $99.16\% \approx 99.2\%$.

(A) For association,the van't Hoff factor $i$ is given by $i = 1 + (\frac{1}{n} - 1) \alpha$.
Given $n = 3$ and $\alpha = 0.80$,we have $i = 1 + (\frac{1}{3} - 1) \times 0.80 = 1 - 0.5333 = 0.4667$.
Depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.
$\Delta T_f = 0.4667 \times 5.12 \times 0.25 = 0.597 \ K \approx 0.6 \ K$.
The freezing point of the solution $T_s = T_f^o - \Delta T_f$.
$T_f^o$ of benzene in Kelvin is $273.15 + 5.5 = 278.65 \ K$.
$T_s = 278.65 - 0.6 = 278.05 \ K$.

(B) Step $1$: Calculate the theoretical depression in freezing point $(\Delta T_{cal})$ assuming no association.
$\Delta T_{cal} = K_f \times m = K_f \times \frac{w \times 1000}{M_w \times W} = 4.9 \times \frac{2 \times 1000}{122 \times 25} = 3.213 \ K$.
Step $2$: Calculate the van't Hoff factor $(i)$.
$i = \frac{\Delta T_{exp}}{\Delta T_{cal}} = \frac{1.62}{3.213} \approx 0.5042$.
Step $3$: Relate $i$ to the degree of association $(\alpha)$ for dimerization $(n=2)$.
$i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
$0.5042 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.4958 \implies \alpha = 0.9916$.
Percentage association = $99.16 \% \approx 99.2 \%$.

(C) For a solute undergoing dissociation,the number of particles increases,which leads to an increase in the observed colligative property. Since $i = \frac{\text{observed colligative property}}{\text{normal colligative property}}$,the value of $i$ is greater than $1$ $(i > 1)$.
In the case of association,the number of particles decreases,leading to a decrease in the observed colligative property. Therefore,the value of $i$ is less than $1$ $(i < 1)$.

(B) The molar mass of phenol $(C_6H_5OH)$ is $M_{Theo} = 94 \ g/mol$.
For dimerization,$n = 2$ and the degree of association $\alpha = 0.60$.
The van't Hoff factor $i$ is given by $i = 1 - \alpha + \frac{\alpha}{n} = 1 - 0.60 + \frac{0.60}{2} = 0.4 + 0.3 = 0.7$.
The relation between observed and theoretical molecular weight is $M_{Obs} = \frac{M_{Theo}}{i}$.
$M_{Obs} = \frac{94}{0.7} \approx 134.28 \ g/mol$.

(B) The van't Hoff factor $i$ is given by the ratio of theoretical molecular weight to observed molecular weight: $i = \frac{M_{th}}{M_{obs}} = \frac{170}{90} \approx 1.889$.
For the dissociation of silver nitrate $(AgNO_3 \rightarrow Ag^+ + NO_3^-)$,the number of ions produced per formula unit is $n = 2$.
The relationship between $i$ and the degree of dissociation $\alpha$ is $i = 1 + (n - 1)\alpha$.
Substituting the values: $1.889 = 1 + (2 - 1)\alpha$.
$1.889 = 1 + \alpha$.
$\alpha = 0.889$.
Therefore,the degree of dissociation is $88.9 \%$.

(C) The formula for freezing point depression is $\Delta T_f = i \times K_f \times m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{20 \ g / 172 \ g/mol}{0.050 \ kg} = \frac{0.11628 \ mol}{0.050 \ kg} = 2.3256 \ mol/kg$.
Given $\Delta T_f = 2 \ K$ and $K_f = 1.72 \ K \ kg \ mol^{-1}$,substitute these values into the equation:
$2 = i \times 1.72 \times 2.3256$.
Solving for $i$:
$i = \frac{2}{1.72 \times 2.3256} = \frac{2}{4} = 0.5$.
Thus,the Van't Hoff factor is $0.5$.

(A) Two solutions are isotonic if their osmotic pressures are equal,i.e.,$\pi_1 = \pi_2$.
For $Na_2SO_4$,the dissociation is $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,so the van't Hoff factor $i = 1 + 2\alpha$.
For glucose (a non-electrolyte),$i = 1$.
Equating the osmotic pressures: $i_1 c_1 RT = i_2 c_2 RT$.
$(1 + 2\alpha) \times 0.004 = 1 \times 0.01$.
$1 + 2\alpha = \frac{0.01}{0.004} = 2.5$.
$2\alpha = 1.5$.
$\alpha = 0.75$.
Therefore,the degree of dissociation is $75 \ \%$.

(C) The molar mass of phenol $(C_6H_5OH)$ is $94 \ g \ mol^{-1}$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{mass of solute (g)}}{\text{molar mass} \times \text{mass of solvent (kg)}} = \frac{20 \ g}{94 \ g \ mol^{-1} \times 1 \ kg} = 0.2128 \ mol \ kg^{-1}$.
Using the freezing point depression formula: $\Delta T_f = i \times K_f \times m$.
$0.69 = i \times 5.12 \times 0.2128$.
$i = \frac{0.69}{1.0895} \approx 0.633$.
For the association of phenol into a dimer $(2C_6H_5OH \rightleftharpoons (C_6H_5OH)_2)$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
$0.633 = 1 - \frac{\alpha}{2}$.
$\frac{\alpha}{2} = 1 - 0.633 = 0.367$.
$\alpha = 0.734$,which is $73.4 \%$. The closest option is $73.3 \%$.

(D) Acetic acid $(CH_3COOH)$ contains a carboxylic group $(-COOH)$ which can form intermolecular hydrogen bonds in non-polar solvents like benzene.
Due to this,acetic acid molecules undergo association to form dimers.
This association reduces the number of effective particles in the solution,which affects the colligative property (freezing point depression) and leads to an abnormal molar mass.

(C) The van't Hoff factor $(i)$ and the degree of association $(\alpha)$ are related by the formula:
$i = 1 - \alpha \left( 1 - \frac{1}{n} \right)$
Given: $i = 0.9$ and $\alpha = 0.2$.
Substituting the values:
$0.9 = 1 - 0.2 \left( 1 - \frac{1}{n} \right)$
$0.2 \left( 1 - \frac{1}{n} \right) = 1 - 0.9$
$0.2 \left( 1 - \frac{1}{n} \right) = 0.1$
$1 - \frac{1}{n} = \frac{0.1}{0.2} = 0.5$
$1 - 0.5 = \frac{1}{n}$
$0.5 = \frac{1}{n}$
$n = \frac{1}{0.5} = 2$
Therefore,the value of $n$ is $2$.

(A) The dimerization reaction is: $2C_6H_5COOH \to (C_6H_5COOH)_2$
Degree of association $\alpha = 0.80$.
Van't Hoff factor $i = 1 - \alpha + \frac{\alpha}{n} = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$.
Depression in freezing point formula: $\Delta T_f = i \times K_f \times m$,where $m$ is molality.
$m = \frac{w \times 1000}{M_{solute} \times W_{solvent(g)}} = \frac{w \times 1000}{122 \times 30}$.
Substituting the values: $2 = 0.6 \times 5 \times \frac{w \times 1000}{122 \times 30}$.
$2 = 3 \times \frac{w \times 100}{122 \times 3} = \frac{w \times 100}{122}$.
$w = \frac{2 \times 122}{100} = 2.44 \ g$.

(A) The van't Hoff factor $i$ is given by the ratio of normal molar mass to observed molar mass:
$i = \frac{\text{Normal molar mass}}{\text{Observed molar mass}} = \frac{164}{65.6} = 2.5$
For the dissociation of $MX_2 \rightarrow M^{2+} + 2X^-$,the number of ions produced per formula unit is $n = 3$.
The relationship between the van't Hoff factor $i$ and the degree of dissociation $\alpha$ is:
$i = 1 + \alpha(n - 1)$
Substituting the values:
$2.5 = 1 + \alpha(3 - 1)$
$2.5 = 1 + 2\alpha$
$1.5 = 2\alpha$
$\alpha = 0.75$
Percentage of ionization $= \alpha \times 100 = 0.75 \times 100 = 75 \%$.
Thus,the correct option is $A$.

(A) The molar mass of naphthoic acid $(C_{11}H_8O_2)$ is calculated as: $(11 \times 12) + (8 \times 1) + (2 \times 16) = 132 + 8 + 32 = 172 \ g \ mol^{-1}$.
The theoretical depression in freezing point $(\Delta T_f)$ is calculated using the formula: $\Delta T_f = K_f \times m$,where $m$ is the molality.
$m = \frac{\text{mass of solute}}{\text{molar mass of solute} \times \text{mass of solvent in kg}} = \frac{20 \ g}{172 \ g \ mol^{-1} \times 0.050 \ kg} = \frac{20}{8.6} \approx 2.325 \ mol \ kg^{-1}$.
$\Delta T_{f, \text{calc}} = 1.72 \ K \ kg \ mol^{-1} \times \frac{20 \times 1000}{172 \times 50} = 1.72 \times \frac{20000}{8600} = 1.72 \times 2.3255 = 4 \ K$.
The Van't Hoff factor $(i)$ is defined as: $i = \frac{\Delta T_{f, \text{obs}}}{\Delta T_{f, \text{calc}}}$.
Given $\Delta T_{f, \text{obs}} = 2 \ K$,we have $i = \frac{2}{4} = 0.5$.

(B) The van't Hoff factor $(i)$ is calculated as: $i = \frac{\text{Calculated molar mass}}{\text{Observed molar mass}} = \frac{170}{92.64} = 1.835$.
For the dissociation of silver nitrate $(AgNO_3 \rightarrow Ag^+ + NO_3^-)$,the relation between $i$ and the degree of dissociation $(\alpha)$ is given by: $i = 1 + (n-1)\alpha$,where $n = 2$.
Substituting the values: $1.835 = 1 + (2-1)\alpha$.
$\alpha = 1.835 - 1 = 0.835$.
Therefore,the degree of dissociation in percentage is $0.835 \times 100 = 83.5 \%$.

(C) For the association reaction: $nA \rightleftharpoons A_n$
Initial moles: $1$ (for $A$) and $0$ (for $A_n$)
Moles at equilibrium: $(1 - \alpha)$ for $A$ and $\frac{\alpha}{n}$ for $A_n$
Total moles at equilibrium $= 1 - \alpha + \frac{\alpha}{n}$
The van't Hoff factor $i$ is defined as the ratio of total moles at equilibrium to the initial moles:
$i = \frac{1 - \alpha + \frac{\alpha}{n}}{1} = 1 - \alpha + \frac{\alpha}{n}$

(D) The reaction is $6 HCHO \rightleftharpoons C_6H_{12}O_6$.
Let the initial concentration of $HCHO$ be $c$.
At equilibrium,the concentration of $HCHO$ is $c(1-\alpha)$ and the concentration of $C_6H_{12}O_6$ is $\frac{c\alpha}{6}$.
The total number of moles at equilibrium is $n_{total} = c(1-\alpha) + \frac{c\alpha}{6} = c(1 - \frac{5\alpha}{6})$.
The van't Hoff factor $i$ is given by $\frac{n_{total}}{n_{initial}} = \frac{c(1 - \frac{5\alpha}{6})}{c} = 1 - \frac{5\alpha}{6}$.
Also,$i = \frac{M_{theoretical}}{M_{observed}} = \frac{30}{150} = 0.2$.
Equating the two expressions: $1 - \frac{5\alpha}{6} = 0.2$.
$\frac{5\alpha}{6} = 0.8$.
$\alpha = \frac{0.8 \times 6}{5} = 0.96$.