Answer to Question #165799 in Chemistry for Phyroe

Question

Calculate the formula weight of a nonelectrolyte, 5 g of which dissolved in 200 g of water gives solution which has a freezing point of negative 1.04 degrees celsius.

Answer 1

Solution:

1) Determine temperature change (∆T):

The freezing point of pure water is 0°C

∆T = 0°C − (-1.04)°C = 1.04°C

2) Determine how many moles of a nonelectrolyte dissolved:

ΔT = i × K_f × m

K_f for water is 1.86°C kg mol¯¹

i = 1 (for nonelectrolyte)

m = the molality of the solute = Moles of nonelectrolyte / kilograms solvent

kilograms solvent = 0.2 kg

Hence,

ΔT = (1) × (1.86°C kg mol¯¹) × (Moles of nonelectrolyte / 0.2 kg) = 1.04°C

Moles of nonelectrolyte / 0.2 = 0.55914

Moles of nonelectrolyte = 0.11183 mol

Moles of nonelectrolyte = Mass of nonelectrolyte / Formula weight of nonelectrolyte

Formula weight of nonelectrolyte = Mass of nonelectrolyte / Moles of nonelectrolyte

Hence,

Formula weight of nonelectrolyte = (5 g) / (0.11183 mol) = 44.7 g/mol

Formula weight of nonelectrolyte = 44.7 g/mol

Answer: The formula weight of a nonelectrolyte is 44.7 g/mol.