18.9 g of CH2ClCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1°C. The van’t Hoff factor for CH2ClCOOH will be (Kf for water = 1.86 K molal–1)

To solve this problem, we need to use the formula for the depression in freezing point, which is ΔTf = i * Kf * m, where ΔTf is the depression in freezing point, i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality.

Step 1: Calculate the molality (m) Molality is the number of moles of solute per kilogram of solvent. First, we need to find the molar mass of CH2ClCOOH. The molar mass of CH2ClCOOH is approximately 94.5 g/mol. So, the number of moles of CH2ClCOOH is 18.9 g / 94.5 g/mol = 0.2 mol. The mass of water is 500 g, which is 0.5 kg. Therefore, the molality is 0.2 mol / 0.5 kg = 0.4 mol/kg.

Step 2: Substitute the values into the formula The depression in the freezing point ΔTf is given as 1°C. The cryoscopic constant Kf for water is given as 1.86 K kg/mol. We substitute these values and the calculated molality into the formula: 1°C = i * 1.86 K kg/mol * 0.4 mol/kg.

Step 3: Solve for i To find the van't Hoff factor i, we rearrange the formula to solve for i: i = 1°C / (1.86 K kg/mol * 0.4 mol/kg) = 1.34.

Therefore, the van't Hoff factor for CH2ClCOOH is approximately 1.34.