A solution of 0.2113 g of water dissolved in 25.0 g of a solvent freezes at 11.5 °C below the freezing point of the solvent. What is Kf for this solvent?

moles H2O = 0.2113/molar mass H2O.

Solve for moles.
molality = moles H2O/kg solvent.
Solve for molality.
delta T = Kf*m
solve for Kf.

To find the Kf (cryoscopic constant) for the solvent, we can use the equation:

ΔTf = Kf * m,

where ΔTf is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution.

In this case, we know that the freezing point depression is 11.5 °C below the freezing point of the solvent. Therefore, ΔTf = -11.5 °C.

We also know the mass of the solvent in the solution is 25.0 g.

To find the molality, we need to calculate the moles of water in the solution. First, we convert the mass of water to moles:

moles of water = mass of water / molar mass of water.

The molar mass of water (H2O) is approximately 18 g/mol.

moles of water = 0.2113 g / 18 g/mol = 0.011739 mol.

Next, we calculate the molality:

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

mass of solvent in kg = 25.0 g / 1000 g/kg = 0.0250 kg.

molality (m) = 0.011739 mol / 0.0250 kg = 0.4695 mol/kg.

Now, we can substitute the values into the equation to solve for Kf:

-11.5 °C = Kf * 0.4695 mol/kg.

Rearranging the equation to solve for Kf:

Kf = ΔTf / m = -11.5 °C / 0.4695 mol/kg.

Kf ≈ -24.5 °C/mol.

Therefore, the cryoscopic constant (Kf) for this solvent is approximately -24.5 °C/mol.