Question

A 0.457 M fructose solution in water has a density of 1.03g/ml. Find the molality of the solution, and the percent by mass and mole fraction of fructose in this solution.

Answer 1

Ans :

Let the volume of solution be 1 L

Molarity = number of mol of solute / volume of solution in L

0.457 = n / 1

n = 0.457 mol

So mass of fructose in solution = mol x molar mass

= 0.457 mol x 180.16 g/mol = 82.33 grams

Density = mass / volume

1.03 = m / 1

mass = 1.03 kg = 1030 grams

Mass of water in solution = 1030 grams - 82.33 grams

= 947.67 g

molality = mol / mass of solvent in kg

= 0.457 / 0.94767

= 0.482 m

percent by mass = ( mass of fructose / mass of solution) x 100

= ( 82.33 / 1030) x 100

= 7.99 %

Number of mol of water = mass / molar mass

= 947.67 / 18.01528

= 52.6 mol

Total number of mol = 0.457 mol + 52.6 mol = 53.06 mol

mole fraction of fructose = mol fructose / total moles

= 0.457 / 53.06

= 0.0086