Question

A solution of fructose (C6H12O6) in water is 18.00 % fructose by mass. Its density is 1.0728 g cm-3 at 20 °C. Compute its molarity, mole fraction, and molality at this temperature.

Answer 1

given

mass % = 18 % this means
mass of fructose = 18 g
total mass = 100 g
density of solution=1.0728 g cm-3
volume of solution = mass/density = 100 g/1.0728 g cm-3
= 93.214 ml = 0.093214 L**

mass of water = 100g -18 g= 82 g
molar mass of water = 18.015 g/mol
number of moles of water = mass/molar mass
= 82 g/18.015 g/mol= 4.55 mols***
molar mass of fructose = 180.16 g/mol
mass of fructose = 18 g
number of moles of fructose = 18 g/180.16 g/mol = 0.0999 mols**
**
molarity = moles of fructose/volume of solution in L
= 0.0999 mols/0.093214 L
=1.072 M
******************

mole fraction of fractose = moles of fractose/total number of moles
=0.0999 mols/0.0999 mol+4.55 mol
= 0.0215
*******************
molality = moles of solute/mass of solvent in Kg
=0.0999 mols/0.082 kg
= 1.218 m
**************