Question

A certain substance X has a normal freezing point of −10.9∘C and a molal freezing point depression constant Xf​=3.83 "C − kg ⋅ mol −1, A solution is prepared by dissolving some urea ((NH2​)2​CO) in 650.8 of X. This solution freeres at −17.2 C. Calculate the mass of urea that was dissolved, Feand your answer to 2 significant digits.

Answer 1

The given parameters are, The freezing point depression constant Xf​ = 3.83 "C − kg ⋅ mol −1Normal freezing point of substance X

= −10.9∘CVolume of X

= 650.8 g

Freezing point of the solution

= −17.2 C

The formula for calculating the freezing point depression (ΔTf) of a solution isΔTf

= Kf × molalitywhere Kf is the molal freezing point depression constant and molality is the molality of the solution.We know that molality (m) of the solution is,

m = (moles of solute) / (mass of solvent in kg)Let's assume the mass of urea (NH2​)2​CO is Fe.Fe has dissolved in the volume X which has a mass of 650.8 g. Molar mass of urea (NH2​)2​CO

= (2 × 14) + 2 + 12 + 16

= 60 g/molNo. of moles of urea, n

= Fe / MWe need to find the mass of urea (Fe) that is dissolved in the solution.To find the molality of the solution, we can use the formula, ΔTf = Kf × molality,ΔTf

= (normal freezing point - freezing point of the solution)

= (-10.9 - (-17.2))

= 6.3 °Cmolality

= ΔTf / Kf

= 6.3 / 3.83

= 1.644 mol/kg

Since 1.644 mol/kg is the molality of the solution. We can write, 1.644

= n / 0.6508Fe

= (1.644 × 60 × 0.6508)

= 63.1 g

Hence, the mass of urea that was dissolved in the solution is 63.1 g.