Mass Percent, Mole Fraction, and Molality 1. A 1.37 M solution of citric acid, H
ID: 1036664 • Letter: M
Question
Mass Percent, Mole Fraction, and Molality1. A 1.37 M solution of citric acid, H3C6H5O7, in water has a density of 1.10 g/mL. Calculate mass percent, molality, and mole fraction of the citric acid.
2. Assume you dissolve 45.0 g of camphor, C10H16O, in 425 mL of ethanol, C2H5OH. Calculate the molality, mol fraction, and mass percent of camphor in this solution. The density of ethanol is 0.785 g/mL.
Homework #3-Mass Percent, Mole Fraction and Molality L. A 1.37 M solution of citric acid, HCaH.O, in water has a density of 1.10 g/mL. Calculate mass percent, molality and mole fraction of the citric acid.22 1.37 2. Assume you dissolve 45.0 g of camphor, CieHisO, in 425 mL of ethanol, Co HOH. Calculate the molality, mole fraction and mass percent of camphor in this solution. The density of ethanol is 0.785 gml. 123 g/mL. Calculate the mass percent and molality of the sulfuric acid
Explanation / Answer
Ans. #1. Let the volume of solution be 1.0 L = 1000.0 mL
# Moles of citric acid = Molarity x Vol. in liters = 1.37 M x 1.0 L = 1.37 mol
Mass of citric acid = Moles x MW = 1.37 M x (192.12532 g/ mol) = 263.212 g
Mass of solution = Volume x Density = 1000.0 mL x (1.10 g/ mL) = 1100.0 g
Mass of solvent (water) = Mass of solution – Mass of solute (citric acid)
= 1100.0 g – 263.212 g
= 836.788 g = 0.836788 kg
# Mass % = (Mass of citric acid / Mass of solution) x 100
= (263.212 g / 1100.0 g) x 100
= 23.93 %
# Molality = Moles of solute / Mass of solvent in kg = 1.37 mol / 0.836788 kg
= 01.637 m
# Moles of water = 836.788 g / 18.0 g mol-1 = 46.488 mol
Total moles = 1.37 mol (citric acid) + 46.488 mol (water) = 47.858 mol
Mole fraction of citric acid = 1.37 mol / 47.858 mol) = 0.0286
#2. Moles of camphor = 45.0 g / (152.23644 g/ mol) = 0.2956 mol
Mass of ethanol = Volume x Density = 425 ml x (0.785 g/ mol)
= 333.625 g = 0.333625 kg
Moles of ethanol = 333.625 g / (46.06904 g/mol) = 7.2418 mol
# Molality = 0.2956 mol / 0.333625 kg = 0.8860 m
# Mole fraction of camphor = 0.2956 mol / (0.2956 + 7.2418) mol = 0.0392
# Mass % = [45.0 g / (45.0 + 333.625) g] x 100 = 11.885 %
#3. Let the volume of solution be 1.0 L = 1000.0 mL
# Moles of H2SO4 = Molarity x Vol. in liters = 3.75 M x 1.0 L = 3.75 mol
Mass of H2SO4 = Moles x MW = 3.75 M x (98.07948 g/ mol) = 367.79805 g
Mass of solution = Volume x Density = 1000.0 mL x (1.23 g/ mL) = 1230.0 g
Mass of solvent (water) = Mass of solution – Mass of solute (citric acid)
= 1230.0 g – 367.79805 g
= 862.20195 g = 0.86220195 kg
# Mass % = (Mass of H2SO4 / Mass of solution) x 100
= (367.79805 g / 1230.0 g) x 100
= 29.90 %
# Molality = Moles of solute / Mass of solvent in kg = 3.75 mol / 0.86220195 kg
= 4.349 m