Question

1.02 g of urea when dissolved in 98.5 g of certain solvent decreases its freezing point by 0.211

K. 1.60 g of unknown compound when dissolved in 86.0 g of the same solvent depresses the

freezing point by 0.34 K. Calculate the molar mass of the unknown compound. (Urea

! , = 14, ! = 12, = 16, = 1) ​

Answer 1

molar mass of unknown compound is 67g/mol

mass of solute = 1.02g

molar mass of solute = 60g/mol

so, no of moles of solute = 1.02/60 = 0.102/6 = 0.017

molality , m = no of moles of solute/mass of solvent in Kg

= 0.017/98.5 × 1000

= 17/98.5

using formula, ∆T_f = k_f × m

⇒0.211 = k_f × 17/98.5

⇒0.211 = k_f × 0.17258

⇒k_f = 0.211/0.17258 = 1.2226

again, mass of unknown solute = 1.6 g

molar mass of solute = M

molality = (1.6/M)/86 × 1000 = 1600/86 M

= 18.6046/M

using formula, ∆T_f = k_f × m

⇒0.34 = 1.2226 × 18.6046/ M

⇒M = 1.2226 × 18.6046/0.34 = 66.89 ≈ 67 g

Answer 2

Wb is equal to 1.02 g wa is equal to 98 .5 g delta Tf is equal to 0.211 k Mb is equal to 60 g mol-1