1 g of benzoic acid dissolved in 12.5 g of benzene shows a depression in freezing point equal to 1.62 K. If benzoic acid forms dimer in solution then percentage association of acid will be (Kf for benzene = 4.9 K kg mol–1
Question
1 g of benzoic acid dissolved in 12.5 g of benzene shows a depression in freezing point equal to 1.62 K. If benzoic acid forms dimer in solution then percentage association of acid will be (Kf for benzene = 4.9 K kg mol–1
Solution
Step 1: Calculate the molality of the solution
The molality (m) of the solution is calculated using the formula:
m = moles of solute / mass of solvent in kg
Given that the mass of benzoic acid (solute) is 1 g and its molar mass is 122 g/mol, the moles of benzoic acid is 1/122 = 0.0082 mol.
The mass of benzene (solvent) is 12.5 g = 0.0125 kg.
So, m = 0.0082 / 0.0125 = 0.656 mol/kg
Step 2: Calculate the theoretical depression in freezing point
The theoretical depression in freezing point (ΔTf) is calculated using the formula:
ΔTf = Kf * m
Given that the cryoscopic constant (Kf) for benzene is 4.9 K kg/mol, the theoretical ΔTf is 4.9 * 0.656 = 3.21 K.
Step 3: Calculate the degree of association
The degree of association (α) is calculated using the formula:
α = (Theoretical ΔTf / Experimental ΔTf) - 1
Given that the experimental ΔTf is 1.62 K, the degree of association is (3.21 / 1.62) - 1 = 0.98 or 98%.
So, the percentage association of benzoic acid is 98%.
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