Answer:
Molality → 0.188 m
Mole fraction of fructose → 0.00337
Mass percent of fructose in solution → 3.29 %
Molarity → 0.183 M
Explanation:
Solute → 34 g of fructose
Solvent → 1000 g of water
Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.
We take account density to calculate, the solution's density
1.0078 g/mL = 1034 g / mL
1034 g / 1.0078 g/mL = 1026 mL
Molal concentration → moles of solute in 1kg of solvent
Moles of fructose = mass of fructose / molar mass
34 g/ 180g/mol = 0.188 mol
0.188 mol/1kg = 0.188 m
Mole fraction of fructose = Moles of fructose / Total moles
We determine the moles of water
Moles of water = 1000 g / 18 g = 55.5 mol
Total moles = moles of fructose + moles of water
0.188 mol + 55.5 mol = 55.743 mol
0.188 mol / 55.743 mol = 0.00337
Mass percent = mass of fructose in 100 g of solution
(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %
Molarity = Moles of solute in 1L of solution
We can also say mmol of solute in 1 mL of solution
0.188 mol of fructose = 188 mmol of fructose
Molarity = 188 mmol / 1026 mL of solution = 0.183 M
