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A 0.200 m K2SO4 solution has a freezing point of −0.53°C. What is the van't Hoff factor for this solution?

Kf = 1.86°C/m

User DShah
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1 Answer

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To solve this problem, we can use the equation for the freezing point depression:

ΔTf = Kf * i * molality

where ΔTf is the change in freezing point, Kf is the freezing point depression constant (given as 1.86°C/m), i is the van't Hoff factor, and molality is the number of moles of solute per kilogram of solvent.

We are given that the freezing point depression is 0.53°C and the molality is:

molality = moles of solute / mass of solvent (in kg)

We don't have the mass of solvent, but we can assume that it is 1 kg (1000 g) since the molality is in moles per kilogram of solvent. This means that the mass of solute (K2SO4) is 0.200 * 1000 = 200 g.

The molar mass of K2SO4 is 174.26 g/mol, so the number of moles of solute is:

moles of solute = 200 g / 174.26 g/mol = 1.147 mol

Substituting these values into the freezing point depression equation, we get:

0.53 = 1.86 * i * (1.147/1)

Solving for i, we get:

i = 0.53 / (1.86 * 1.147) = 0.247

Therefore, the van't Hoff factor for this solution is approximately 0.247
User Khanmizan
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