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The normal freezing point of a certain liquid Y is -8.70°C, but when 24.17g of urea (NH2)2CO are dissolved in 550 g of Y, it is found that the solution freezes at -13.7°C instead. Use this information to calculate the molal freezing point depression constant Kf of substance Y.

Respuesta :

baraltoa

Answer:

6.8 ÂșC m⁻Âč

Explanation:

The strategy here is to use the equation for the freezing point depression

ΔTf = Kf x m   ⇒   Kf = ΔTf / m

where  k is the freezing point depression constant,ΔTf  is the change in the freezing point of the solution ( freezing point pure solvent minus freezing point of the solution), and m is the molality, mol of solute per kilogram solvent.

Thus

ΔTf (ÂșC) = -8.7 ÂșC - ( -13.7 )C  =  5.0 ÂșC

mol (NH₂)₂CO = mass / molar mass = 24.17 g / 60.06 gmol⁻Âč = 0.40 mol

m = 0.40 mol / 0.550 Kg = 0.73 m

Kf = ΔTf / m =  5.0 ÂșC / 0.73 m = 6.8 ÂșCm⁻Âč

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