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A 6.0 gram sample of an unknown compound is dissolved in 125 g of water. The solutions freezing point is lowered to 1.38 C° below the freezing point of pure water. What is the molar mass of the sample? K for water is 1.86 C°/m.

Respuesta :

Edufirst
Change in the freezing point = Kf * m

m = molality
Kf = 1.86 °C/m

m = Change in the freezing point / k = 1.38°C / 1.86 °C/m =0.742 m

m = #mol of solute / Kg of solvent

Kg of solvent = 125g/1000 g/kg = 0.125 kg.

#mol of solute = m* Kg of solvent = 0.742m * 0.125 kg =0.09275mol

Molar mass = 6. 0 g / 0.09275 mol = 64.7 g/mol

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