1.00 g of a non-electrolyte solute dissolved in 50 g of benzene, freezing point lowered by 0.40 K. The freezing point depression constant of benzene is 5.12 K, kg mol^−1 . Find the molar mass of the solute.
Question
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene, freezing point lowered by 0.40 K. The freezing point depression constant of benzene is 5.12 K, kg mol^−1 . Find the molar mass of the solute.
Solution
To find the molar mass of the solute, we can use the formula for freezing point depression:
ΔTf = Kf * m
where: ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.
First, we need to find the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. We can rearrange the formula to solve for molality:
m = ΔTf / Kf
Substituting the given values:
m = 0.40 K / 5.12 K kg mol^-1 = 0.078125 mol/kg
Next, we need to convert the mass of the solvent (benzene) from grams to kilograms:
50 g = 0.05 kg
Now we can find the number of moles of solute:
number of moles = molality * mass of solvent = 0.078125 mol/kg * 0.05 kg = 0.00390625 mol
Finally, we can find the molar mass of the solute by dividing the mass of the solute by the number of moles:
molar mass = mass of solute / number of moles = 1.00 g / 0.00390625 mol = 256 g/mol
So, the molar mass of the solute is 256 g/mol.
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