Phenol (C6H5OH) associates in water forming double molecules. When 0.658 g of phenol is dissolved in 35 g of water it depresses the freezing point by 0.215°C. Calculate degree of association of phenol in water. Molal depression freezing point constant of water = 1.85°C kg mol–1.
Question
Phenol (C6H5OH) associates in water forming double molecules. When 0.658 g of phenol is dissolved in 35 g of water it depresses the freezing point by 0.215°C. Calculate degree of association of phenol in water. Molal depression freezing point constant of water = 1.85°C kg mol–1.
Solution
Sure, here are the steps to calculate the degree of association of phenol in water:
Step 1: Calculate the molality of the solution. Molality (m) = moles of solute / mass of solvent in kg First, calculate the moles of phenol. The molar mass of phenol (C6H5OH) is approximately 94 g/mol. So, moles of phenol = 0.658 g / 94 g/mol = 0.007 mol The mass of water is 35 g, which is 0.035 kg. So, molality (m) = 0.007 mol / 0.035 kg = 0.2 mol/kg
Step 2: Calculate the experimental molal depression constant (Kf) Kf = ΔTf / m Where ΔTf is the depression in freezing point, which is 0.215°C. So, Kf = 0.215°C / 0.2 mol/kg = 1.075°C kg/mol
Step 3: Calculate the degree of association (α) The theoretical Kf of water is given as 1.85°C kg/mol. The degree of association (α) is the ratio of experimental Kf to theoretical Kf. So, α = 1.075°C kg/mol / 1.85°C kg/mol = 0.58
So, the degree of association of phenol in water is 0.58 or 58%.
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