Abnormal molecular mass Questions in English

(D) Given:
Mass of solute $(w_B)$ = $2.5 \ g$
Mass of solvent $(w_A)$ = $100 \ g$
Elevation in boiling point $(\Delta T_b)$ = $0.3 \ K$
$K_b = 0.52 \ K \ kg \ mol^{-1}$
Degree of association $(\alpha)$ = $0.8$
For dimerization: $2X \rightleftharpoons X_2$
Van't Hoff factor $(i)$ = $1 - \alpha + \frac{\alpha}{2} = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$
Formula for elevation in boiling point:
$\Delta T_b = i \times K_b \times m$
$\Delta T_b = i \times K_b \times \left( \frac{w_B \times 1000}{M_B \times w_A} \right)$
$0.3 = 0.6 \times 0.52 \times \left( \frac{2.5 \times 1000}{M_B \times 100} \right)$
$0.3 = 0.6 \times 0.52 \times \left( \frac{25}{M_B} \right)$
$M_B = \frac{0.6 \times 0.52 \times 25}{0.3}$
$M_B = 2 \times 0.52 \times 25 = 26 \ g \ mol^{-1}$

(C) Given: Degree of association $\alpha = 80\% = 0.8$ and $n = 4$ (for tetramerization).
Van't Hoff factor $i = 1 + (\frac{1}{n} - 1)\alpha = 1 + (\frac{1}{4} - 1) \times 0.8 = 1 - 0.6 = 0.4$.
Depression in freezing point $\Delta T_f = i \times K_f \times m$.
$0.3 = 0.4 \times 1.86 \times \frac{2.5 \times 1000}{M_A \times 100}$.
$0.3 = 0.4 \times 1.86 \times \frac{25}{M_A}$.
$M_A = \frac{0.4 \times 1.86 \times 25}{0.3} = \frac{18.6}{0.3} = 62 \text{ g mol}^{-1}$.

(A) $1$. Calculate the observed molality $(m_{obs})$: $\Delta T_f = K_f \times m_{obs} \implies 2 = 5 \times m_{obs} \implies m_{obs} = 0.4 \ mol \ kg^{-1}$.
$2$. Calculate the theoretical molality $(m_{theo})$: Moles of benzoic acid $= \frac{2.44 \ g}{122 \ g \ mol^{-1}} = 0.02 \ mol$. Mass of solvent $= 0.03 \ kg$. $m_{theo} = \frac{0.02}{0.03} = 0.667 \ mol \ kg^{-1}$.
$3$. Calculate the van't Hoff factor $(i)$: $i = \frac{m_{obs}}{m_{theo}} = \frac{0.4}{0.667} = 0.6$.
$4$. For dimerisation,$i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.
$5$. $0.6 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.4 \implies \alpha = 0.8$.
$6$. Percentage of association $= 0.8 \times 100 = 80\%$.

(A) For dimerization of benzoic acid: $2C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$
Van't Hoff factor $(i) = 1 - \alpha + \frac{\alpha}{n}$
Given $\alpha = 88 \% = 0.88$ and $n = 2$
$i = 1 - 0.88 + \frac{0.88}{2} = 1 - 0.88 + 0.44 = 0.56$
We know,$\Delta T_f = i \cdot K_f \cdot m$
$\Delta T_f = i \cdot K_f \cdot \frac{w_2 \cdot 1000}{M_2 \cdot w_1}$
$1.12 = 0.56 \cdot 4.9 \cdot \frac{x \cdot 1000}{122 \cdot 49}$
$1.12 = 0.56 \cdot 0.1 \cdot \frac{1000x}{122}$
$1.12 = \frac{56x}{122}$
$x = \frac{1.12 \cdot 122}{56} = 2.44 \ g$

(B) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
Mass of acetic acid = $density \times volume = 1.06 \ g \ cm^{-3} \times 1.2 \ mL = 1.272 \ g$.
Moles of acetic acid = $\frac{1.272 \ g}{60 \ g \ mol^{-1}} = 0.0212 \ mol$.
Since $1 \ L$ of water is used,the mass of the solvent is $1000 \ g = 1 \ kg$.
Molality $(m) = \frac{0.0212 \ mol}{1 \ kg} = 0.0212 \ mol \ kg^{-1}$.
Substituting the values into the formula: $0.041 = i \times 1.86 \times 0.0212$.
$i = \frac{0.041}{1.86 \times 0.0212} \approx \frac{0.041}{0.039432} \approx 1.04$.
Thus,the van't Hoff factor is $1.04$.
Therefore,option $(B)$ is correct.

(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi)$.
For $NaCl$ solution: $\pi_1 = i \times C_1 \times R \times T$.
For urea solution: $\pi_2 = C_2 \times R \times T$ (since $i = 1$ for urea).
Given $1.1 \ w \%$ urea solution means $1.1 \ g$ of urea in $100 \ mL$ of solution, so $C_2 = \frac{1.1 \ g / 60 \ g \ mol^{-1}}{0.1 \ L} = 0.1833 \ M$.
Equating the osmotic pressures: $i \times 0.1 = 0.1833$.
$i = 1.83$.
For $NaCl \rightarrow Na^+ + Cl^-$, the number of ions $n = 2$.
The degree of ionization $\alpha$ is given by $\alpha = \frac{i - 1}{n - 1}$.
$\alpha = \frac{1.83 - 1}{2 - 1} = 0.83$.

(B) The osmotic pressure formula is given by $\pi = iCRT$.
Given: $\pi = 4.82 \ atm$,$C = 0.1 \ M$ (for $NaCl$,$N = M$),$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.
Calculating the van't Hoff factor $(i)$:
$i = \frac{\pi}{CRT} = \frac{4.82}{0.1 \times 0.082 \times 300} = \frac{4.82}{2.46} \approx 1.959 \approx 1.96$.
For $NaCl$,the dissociation is $NaCl \rightarrow Na^+ + Cl^-$,so $n = 2$.
The degree of dissociation $(\alpha)$ is related to $i$ by $i = 1 + \alpha(n - 1)$.
$1.96 = 1 + \alpha(2 - 1) \Rightarrow \alpha = 0.96$.
Since $\alpha = x \times 10^{-2}$,we have $0.96 = x \times 10^{-2}$,which gives $x = 96$.

(B) Given: $W_B = 2.44 \ g$,$K_f = 5.0 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 2.2 \ K$.
First,calculate the observed molar mass $(M_{obs})$ using the formula: $\Delta T_f = K_f \times \frac{W_B}{M_{obs}} \times \frac{1000}{W_A}$.
$M_{obs} = \frac{5.0 \times 2.44 \times 1000}{2.2 \times 25} = 221.8 \ g \ mol^{-1}$.
The theoretical molar mass of benzoic acid $(C_6H_5COOH)$ is $122 \ g \ mol^{-1}$.
For association: $2 C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$.
Van't Hoff factor $i = \frac{M_{theoretical}}{M_{observed}} = \frac{122}{221.8} \approx 0.55$.
Also,$i = 1 + (\frac{1}{n} - 1)x$,where $n=2$ for dimer and $x$ is the degree of association.
$0.55 = 1 + (\frac{1}{2} - 1)x \implies 0.55 = 1 - 0.5x$.
$0.5x = 0.45 \implies x = 0.90$.
Percentage association = $90\%$.

(D) The van't Hoff factor $i$ is defined as the ratio of the theoretical molar mass to the observed molar mass:
$i = \frac{(\text{Molar mass})_{\text{theoretical}}}{(\text{Molar mass})_{\text{observed}}} = \frac{142}{50.0} = 2.84$
$Na_2SO_4$ dissociates in water as:
$Na_2SO_4 \rightleftharpoons 2Na^+ + SO_4^{2-}$
Here,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 given by:
$i = 1 + (n-1)\alpha$
Substituting the values:
$2.84 = 1 + (3-1)\alpha$
$2.84 - 1 = 2\alpha$
$1.84 = 2\alpha$
$\alpha = \frac{1.84}{2} = 0.92$

(D) Colligative properties are inversely proportional to the molar mass of the solute $(M \propto \frac{1}{\text{Colligative Property}})$.
When a solute associates in a solvent,the number of particles decreases,which leads to a decrease in the observed colligative property (like boiling point elevation).
Since the observed colligative property is lower than the theoretical value,the experimentally calculated molar mass will be higher than the actual molar mass.

(A) The van't Hoff factor $(i)$ is defined as the ratio of the theoretical (normal) molecular mass to the observed (experimental) molecular mass:
$i = \frac{\text{Molecular mass (theoretical)}}{\text{Molecular mass (experimental)}}$
Given that the experimental molecular mass is double the actual (theoretical) value,we have:
$i = \frac{1}{2} = 0.5$
For the association of a solute to form a dimer,the relationship between the van't Hoff factor $(i)$ and the degree of association $(\alpha)$ is given by:
$i = 1 - \alpha + \frac{\alpha}{n}$
Since it is a dimer,$n = 2$:
$0.5 = 1 - \alpha + \frac{\alpha}{2}$
$0.5 = 1 - \frac{\alpha}{2}$
$\frac{\alpha}{2} = 1 - 0.5 = 0.5$
$\alpha = 1$
Thus,the degree of association is $1$ (or $100\%$).

(D) For the association reaction: $2 A \rightleftharpoons (A)_2$
Given degree of association,$\alpha = 70 \% = 0.70$.
The number of particles associating,$n = 2$.
The van't-Hoff factor $(i)$ is given by the formula: $i = 1 - \alpha + \frac{\alpha}{n}$.
Substituting the values: $i = 1 - 0.70 + \frac{0.70}{2}$.
$i = 0.30 + 0.35 = 0.65$.

(C) The van't Hoff factor $(i)$ is defined as the ratio of the observed colligative property to the calculated (theoretical) colligative property.
$i = \frac{\Delta T_f \text{ (observed)}}{\Delta T_f \text{ (calculated)}}$
Given:
$\Delta T_f \text{ (observed)} = 0.025 \ K$
$i = 2.0$
Substituting the values:
$2.0 = \frac{0.025}{\Delta T_f \text{ (calculated)}}$
$\Delta T_f \text{ (calculated)} = \frac{0.025}{2.0} = 0.0125 \ K$

(C) $BaCl_2$ is an electrolyte that dissociates in water as $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$.
Since the number of particles increases due to dissociation,the observed colligative property (elevation in boiling point) is higher than the theoretical value.
Because the molar mass is inversely proportional to the colligative property,the experimental molar mass is found to be less than the calculated (theoretical) molar mass.

(A) Given:
Mass of solute $(w_2)$ = $20 \ g$
Mass of solvent $(w_1)$ = $50 \ g$
Freezing point depression $(\Delta T_f)$ = $2 \ K$
$K_f$ of benzene = $1.72 \ K \ kg \ mol^{-1}$
Molar mass of naphthoic acid $(C_{11}H_8O_2)$ = $(11 \times 12) + (8 \times 1) + (2 \times 16) = 172 \ g \ mol^{-1}$
Using the formula:
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = i \times K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$
$2 = i \times 1.72 \times \frac{20 \times 1000}{172 \times 50}$
$2 = i \times 1.72 \times \frac{20000}{8600}$
$2 = i \times 1.72 \times 2.3255...$
$2 = i \times 4$
$i = \frac{2}{4} = 0.5$

(B) $\Delta T_{f} = i \times K_{f} \times m$
$\therefore i = \frac{\Delta T_{f}}{K_{f} \times m} = \frac{0.19}{1.86 \times 0.1} = 1.0215 \approx 1.02$
Again from,$\alpha = \frac{i-1}{n-1} = \frac{1.02-1}{2-1} = 0.02$
For $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$
$K_{a} = C \alpha^{2}$
$= 0.1 \times (0.02)^{2} = 0.1 \times 4 \times 10^{-4} = 4 \times 10^{-5}$