Question

a) When 1.13 g of an unknown non-electrolyte is dissolved in 50.0 g of carbon tetrachloride, the freezing point decreased by 5.46 degrees C. If the K_fp of the solvent is 29.8 K/m, calculate the molar mass of the unknown solute.

b) How many moles of solute particles are present in 4.78 mL of 0.304 M (NH₄)₂SO₄?

c) A 0.97 mass % aqueous solution of urea (CO(NH₂)₂) has a density of 1.12 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.

Answer 1

Mass of solute =1.13 g let the molar mass= M

Moles of solute= 1.13/M

Mass of CCl4 solvent =50 gm= 50/1000= 0.05 kg

Molality = moles of solute dissolved in 1 kg of solvent

0.05 kg dissolves 1.13/ M moles of solute

1 kg dissolved 1.13/0.05M=22.6/M

depression in freezing point = kf*molality kf= 29.8 K/m

5.46= 29.8*22.6/M

M= 29.8*22.6/5.46= 123.349 g/mole

b) Solute is (NH4)2SO4 moles of solute in 4.78ml = Molarity* volume of solution= 0.304*4.78/1000 g/mole=0.001453 moles

c) Molecular weight of Urea = 12+16+32=60

basis : 100 gms of solution amount of urea in 100 gms =0.97 gm moles of urea= 0.97/60= 0.016167

Density of the solution =1.12 g/ml

Volume of solution= mass/ density= 100/1.12 ml =89.3 ml =0.089 L

Molarity = moles of Urea/ liter of solution= 0.016167/0.089 M= 0.181 M