Question

If 2 g of benzoic acid is dissolved in 25 g of benzene, its depression in freezing point was found to be 1.62 K. Molal depression constant of benzene is 4.9 K kg mol^-1. If benzoic acid forms a dimer in this solution, what will be the percentage association of benzoic acid? (5)
OR
A 200 cm³ aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of this solution at 300 K is \(2.57\times10^{-3}\) bar. Calculate the molar mass of this protein. (5)

Hint:

Option 1: find the van't Hoff factor i from \(\Delta T_f = iK_f m\), then use \(i = 1 - \alpha/2\) for dimerisation. Option 2: use \(M = wRT/\pi V\) with R = 0.083 L bar K^-1mol^-1.

Answer 1

Option 1 (Benzoic acid dimer):
Step 1 (Concept): Depression in freezing point is a colligative property, \(\Delta T_f = i\,K_f\,m\), where \(i\) is the van't Hoff factor and \(m\) is molality. When solute molecules associate (form a dimer), \(i<1\).
Step 2 (Molar mass of benzoic acid): \(C_6H_5COOH\) has molar mass \(= 122\) g mol^-1.
Step 3 (Molality): moles \(= \dfrac{2}{122} = 0.01639\) mol; molality \(m = \dfrac{0.01639}{0.025\text{ kg}} = 0.6557\) mol kg^-1.
Step 4 (Theoretical depression, no association): \(\Delta T_f(\text{calc}) = K_f\,m = 4.9 \times 0.6557 = 3.213\) K.
Step 5 (van't Hoff factor): \(i = \dfrac{\Delta T_f(\text{obs})}{\Delta T_f(\text{calc})} = \dfrac{1.62}{3.213} = 0.504\).
Step 6 (Association relation): For dimerisation \(2A \rightarrow A_2\), if \(\alpha\) is the degree of association, \(i = 1 - \dfrac{\alpha}{2}\). So \(\alpha = 2(1-i) = 2(1-0.504) = 0.992\).
Step 7 (Percentage association): \[\boxed{\alpha = 99.2\%}\]

Option 2 (Molar mass of protein):
Step 1 (Concept): Osmotic pressure \(\pi = \dfrac{n}{V}RT = \dfrac{w}{M\,V}RT\), so \(M = \dfrac{wRT}{\pi V}\).
Step 2 (Data): \(w = 1.26\) g, \(V = 200\text{ cm}^3 = 0.200\) L, \(R = 0.083\) L bar K^-1 mol^-1, \(T = 300\) K, \(\pi = 2.57\times10^{-3}\) bar.
Step 3 (Substitution): \(M = \dfrac{1.26 \times 0.083 \times 300}{2.57\times10^{-3} \times 0.200}\).
Step 4 (Arithmetic): Numerator \(= 31.374\); Denominator \(= 5.14\times10^{-4}\); \(M = 61039\) g mol^-1.
Step 5 (Result): \[\boxed{M \approx 6.1\times10^{4}\text{ g mol}^{-1}}\]