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3 years ago

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The freezing point of 42.19 g of a pure solvent is measured to be 43.17 ºC. When 2.03 g of an unknown solute (assume the van 't

Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 40.32 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.67 ºC·kg solvent/mol solute).What is the molality of the solution? m
How many moles of solute are present? mol
What is the molecular weight of the solute? g/molIf the freezing point depression constant of the pure solvent is 7.66 ºC·kg solvent/mol solute calculate the molar mass of the solute. g/mol

1 answer:

vovikov84 [41]3 years ago

4 0

Answer:

The molality of the solution is 0.3716 mol/kg

The number of moles of solute is 0.0157 mol

The molecular weight of the solute is 129.30 g/mol

The molar mass of the solute is 129.32 g/mol

Explanation:

m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg

Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol

Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol

When Kf = 7.66 °C.kg/mol

Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol

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