Module in General Chemistry 2
2nd Semester: Quarter 3, Week 3, M –3
Molality - Problem-solving
Calculating Molality
1. What is the molality of a solution containing 20 grams of sodium chloride (NaCl) in 500 grams
of water?
Solution:
1. Molar mass of NaCl
Na = 22.99 g
Cl = 35.45 g
58.44 g/mol
2. Moles of NaCl = mass / molar mass
Mass of solute 20 g NaCl
Moles of NaCl = = = 0.3421 mol
Molar mass of NaCl 58.45 g /mol
1.0 g
3. Mass of solvent (kg) = 500 g
mL
1.0 g
= 500 g X = 0.5 kg
1000 mL
M oles of Solute 0.3421 mol
4. Molality (m) = = = 0.6842 m
kg of solvent 0.5 k g
Example 2: Calculating Mass of Solute
A solution has a molality of 2.5 m. If the mass of the solvent is 200 grams, what is the mass of
the solute (sucrose, C12H22O11)?
Solution:
1. Molality (m) = 2.5 m
1.0 g
2. Mass of solvent (kg) = 2 00 g
mL
1.0 g
= 200 g X = 0.2 kg
1000 mL
3. Moles = m x mass of solvent (kg)
Moles = 2.5 m x 0.2 kg = 0.5 mol
4. Molar mass of sucrose
C = 12 X 12.01 = 144.12
H = 12 X 1.01 = 12.12
O = 12 X 16.00 = 192.00
348.24 g/mol
5. Mass of solute = moles x molar mass
Mass of solute = 0.5 mol x 342.3 g/mol = 171.15 g
Example 3: Calculating Volume of Solution
What volume of a 1.2 m solution of sodium nitrate (NaNO3) contains 0.6 moles of NaNO3?
Solution:
1. Molality (m) = 1.2 m
2. Moles = 0.6 mol
3. Mass of solvent (kg) = moles / m = 0.6 mol / 1.2 m = 0.5 kg
4. Volume of solution (L) = mass of solvent (kg) / density of solvent
5. Assuming density of solvent = 1 g/mL, volume = 0.5 kg x 1000 mL/kg = 500 mL
Example 4: Converting Molality to Molarity
A solution has a molality of 3.5 m. If the density of the solution is 1.2 g/mL, what is its molarity?
Solution:
1. Molality (m) = 3.5 m
2. Density = 1.2 g/mL
3. Mass of solvent (kg) = 1000 g / 1000 = 1 kg
4. Moles = m x mass of solvent (kg) = 3.5 m x 1 kg = 3.5 mol
5. Volume of solution (L) = mass of solution / density = 1200 g / 1.2 g/mL = 1 L
6. Molarity (M) = moles / volume (L) = 3.5 mol / 1 L = 3.5 M
Example 5: Calculating Molality from Molarity
A solution has a molarity of 2.8 M. If the density of the solution is 1.1 g/mL, what is its molality?
Solution:
1. Molarity (M) = 2.8 M
2. Density = 1.1 g/mL
3. Volume of solution (L) = 1000 mL / 1000 = 1 L
4. Moles = M x volume (L) = 2.8 M x 1 L = 2.8 mol
5. Mass of solvent (kg) = (1000 mL x 1.1 g/mL) / 1000 = 1.1 kg
6. Molality (m) = moles / mass of solvent (kg) = 2.8 mol / 1.1 kg = 2.55 m
Example 6: Mixing Solutions
Two solutions are mixed: 200 grams of a 2 m solution of sodium chloride (NaCl) and 300 grams
of a 1 m solution of NaCl. What is the resulting molality?
Solution:
1. Moles of NaCl in first solution = m x mass of solvent (kg) = 2 m x 0.2 kg = 0.4 mol
2. Moles of NaCl in second solution = m x mass of solvent (kg) = 1 m x 0.3 kg = 0.3 mol
3. Total moles = 0.4 mol + 0.3 mol = 0.7 mol
4. Total mass of solvent (kg) = 0.2 kg + 0.3
