Question

Calculate the freezing point of 12.40 g of urea, (NH₂)₂CO, are dissolved in 98.89 g of water. The K_f for water is 1.86 ^oC/m.

Answer 1

Molar mass of (NH2)2CO

=> 2 * (14 + 2 * 1) + 12 + 16

=> 32 + 12 + 16

=> 60 gm/mol

Number of moles of Urea = mass/molar mass = 12.40/60 = 0.2067 moles

Urea is not an ionic compound, hence the vont hoff factor for urea is equal to 1

molality = number of moles of solute/mass of solvent (in Kg)

=> 0.2067/(98.89/1000) = 2.090

Change in Temperature = vont hoff factor * Kf * molality

=> 1 * 1.86 * 2.090

=> 3.887

Hence the depression in freezing point is equal to 3.887C