19.5 g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation

To calculate the van't Hoff factor and dissociation, we need to follow these steps:

1. First, we need to calculate the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. The molar mass of CH2FCOOH is approximately 96.5 g/mol. So, the number of moles of CH2FCOOH is 19.5 g / 96.5 g/mol = 0.202 mol. The mass of water is 500 g, which is 0.5 kg. Therefore, the molality of the solution is 0.202 mol / 0.5 kg = 0.404 m.

2. The formula for the depression in freezing point is ΔTf = Kf * m * i, where ΔTf is the depression in freezing point, Kf is the cryoscopic constant of water (1.86 C/m), m is the molality of the solution, and i is the van't Hoff factor. We can rearrange this formula to solve for i: i = ΔTf / (Kf * m). Substituting the given and calculated values, we get i = 1.0 C / (1.86 C/m * 0.404 m) = 1.34.

3. The van't Hoff factor (i) is a measure of the extent of dissociation of the solute. A value of i greater than 1 indicates that the solute dissociates in solution. In this case, the calculated i value of 1.34 suggests that CH2FCOOH partially dissociates in water.

4. The degree of dissociation (α) can be calculated using the formula α = (i - 1) / (n - 1), where n is the number of ions the solute dissociates into. For CH2FCOOH, n = 2 (it dissociates into CH2FCOO- and H+). Substituting the values, we get α = (1.34 - 1) / (2 - 1) = 0.34 or 34%.

So, the van't Hoff factor is 1.34 and the degree of dissociation is 34%.