if 5.13g of aq solution of sugar c12h22o11 is present in 250 ml of h2o then calculate molality of the solution

Answer 1

The formula for molality is :

moles of solute / kg of solvent

Sugar is the solute, and it has a formula weight of 342.3 g/mol

5.13 g of sugar is 0.0150 moles.

Water is the solvent. Without extra information, I'll assume a density of 1 g/ml.

Therefore 250 ml = 250 g = 0.25 kg

Molality is moles solute/kg of solvent, so 0.0150 moles solute / 0.25 kg solvent = .0599 m = 0.060 m