The solution containing 6.8g of non ionic solute in 100g water was found to freeze at -0.93℃ . If kf for water is 1.86 the molar mass of solute is

Question

I need the answers immediately please help me

Answer 1

This is a freezing point depression problem and uses the formula ∆T = imK

∆T = change in freezing point =  0.93 degrees

i = van't Hoff factor = 1 for a non ionic solute

m = molality = moles solute/kg solvent = ?

K = freezing constant = 1.86º/m

0.93 = (1)(m)(1.86)

m = 0.5 moles solute/kg solvent

Since there is 100 g water (solvent), this is equal to 0.1 kg solvent, so you have...

0.5 moles/kg x 0.1 kg = 0.05 moles of solute

This is equal to the 6.8 g, so you now have

6.8 g/0.05 moles = 136 g/mole = molar mass of the solute