Question

Among the following statements: (A) 1 M NaCl has higher freezing point than 1M glucose solution (B) 1 M NaCl solution has almost same boiling point as 1 M KCl solution, assuming 100% ionisation (C) Apparent molecular weight of NaCl will be less than 58.5 in water because it undergoes dissociation (D) When acetic acid is dissolved in benzene, i = 0 The correct statements are

• A. A. A and B only
• B. B. B and C only
• C. C. A, B and C only
• D. D. A, B, C and D

Answer 1

Answer: (B)

B. B and C only

Answer 2

The problem requires evaluating statements related to colligative properties, which depend on the number of solute particles in a solution. The van't Hoff factor (i) accounts for the dissociation or association of solute particles.

The relevant formulas are:

1. Depression in freezing point: $\Delta T_f = i K_f m$ The freezing point of the solution is $T_f = T_f^0 - \Delta T_f$. A higher $\Delta T_f$ means a lower freezing point.
2. Elevation in boiling point: $\Delta T_b = i K_b m$ The boiling point of the solution is $T_b = T_b^0 + \Delta T_b$. A higher $\Delta T_b$ means a higher boiling point.
3. Apparent Molecular Weight: $i = \frac{\text{Normal Molecular Weight}}{\text{Apparent Molecular Weight}}$

Let's analyze each statement:

(A) 1 M NaCl has higher freezing point than 1M glucose solution

• NaCl: Being a strong electrolyte, NaCl dissociates into Na$^+$ and Cl$^-$ ions. Assuming 100% dissociation, $i_{NaCl} = 2$.
• Glucose (C$_6$H$_{12}$O$_6$): Glucose is a non-electrolyte and does not dissociate. So, $i_{glucose} = 1$.
• Both solutions have the same molarity (1 M), which implies approximately the same molality (m) for dilute aqueous solutions.
• $\Delta T_f (\text{NaCl}) = i_{NaCl} K_f m = 2 K_f m$
• $\Delta T_f (\text{Glucose}) = i_{glucose} K_f m = 1 K_f m$
• Since $2 K_f m > 1 K_f m$, $\Delta T_f (\text{NaCl}) > \Delta T_f (\text{Glucose})$.
• A greater depression in freezing point means a lower freezing point. Therefore, the freezing point of 1 M NaCl solution will be lower than that of 1 M glucose solution.
• Thus, statement (A) is Incorrect.

(B) 1 M NaCl solution has almost same boiling point as 1 M KCl solution, assuming 100% ionisation

• NaCl: Dissociates into Na$^+$ and Cl$^-$. $i_{NaCl} = 2$.
• KCl: Dissociates into K$^+$ and Cl$^-$. $i_{KCl} = 2$.
• Both solutions have the same molarity (1 M), implying approximately the same molality (m).
• The solvent is water, so $K_b$ is the same for both.
• $\Delta T_b (\text{NaCl}) = i_{NaCl} K_b m = 2 K_b m$
• $\Delta T_b (\text{KCl}) = i_{KCl} K_b m = 2 K_b m$
• Since the van't Hoff factors and molalities are the same, their elevations in boiling point ($\Delta T_b$) will be almost identical. Consequently, their boiling points will be almost the same.
• Thus, statement (B) is Correct.

(C) Apparent molecular weight of NaCl will be less than 58.5 in water because it undergoes dissociation

• The normal molecular weight of NaCl is 58.5 g/mol.
• When NaCl dissolves in water, it dissociates into Na$^+$ and Cl$^-$, so $i = 2$.
• The relationship between normal molecular weight ($M_{normal}$) and apparent molecular weight ($M_{apparent}$) is $i = \frac{M_{normal}}{M_{apparent}}$.
• Rearranging, $M_{apparent} = \frac{M_{normal}}{i}$.
• For NaCl, $M_{apparent} = \frac{58.5}{2} = 29.25$ g/mol.
• Since 29.25 g/mol < 58.5 g/mol, the apparent molecular weight of NaCl in water is less than its normal molecular weight. This is a direct consequence of dissociation, which increases the number of particles in solution.
• Thus, statement (C) is Correct.

(D) When acetic acid is dissolved in benzene, i = 0

• Acetic acid (CH$_3$COOH) is known to undergo dimerization in non-polar solvents like benzene due to intermolecular hydrogen bonding. $2 \text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2$
• Dimerization is an association process, where two molecules combine to form one particle. This leads to a decrease in the number of particles in the solution.
• For association, the van't Hoff factor (i) is less than 1. If 100% dimerization occurs, $i = 1/2 = 0.5$.
• An 'i' value of 0 would imply no solute particles or no colligative effect, which is incorrect.
• Thus, statement (D) is Incorrect.

Based on the analysis, statements (B) and (C) are correct.