Exercise | Q 12 | Page 46
An aqueous solution of a certain organic compound has a density of 1.063 g mL-1 , osmotic pressure of 12.16 atm at 25 °C and a freezing point of 1.03 °C. What is the molar mass of the compound?
Solution
Given: Density of a solution = d = 1.063 g mL-1
Osmotic pressure of solution = π = 12.16 atm
Temperature = T = 25 °C = 298.15 K
Freezing point of solution = Tf = - 1.03 °C
To find: Molar mass of a compound
Formulae: 1.
2. π = MRT
3. m =
Calculation: R = 0.08205 dm3 atm K-1 mol-1
= 0 °C - (- 1.03 °C) = 1.03 °C = 1.03 K
Kf of water = 1.86 K kg mol–1
Using formula (i),
m = = 0.554 mol kg-1 = 0.554 m
Using formula (ii),
π = MRT
M = = 0.497 mol dm-3 = 0.497 M
Mass of solvent = = 0.897 kg = 897 g units
Mass of solution = 1.063 g mL-1 × 1000 mL = 1063 g
Mass of solute = 1063 g – 897 g = 166 g
Now, using formula (iii),
m =
∴
= 334 g mol-1
The molar mass of the compound is 334 g mol-1.
Given: Density of a solution = d = 1.063 g mL-1
Osmotic pressure of solution = π = 12.16 atm
Temperature = T = 25 °C = 298.15 K
Freezing point of solution = Tf = - 1.03 °C
To find: Molar mass of a compound
Formulae: 1.
2. π = MRT
3. m =
Calculation: R = 0.08205 dm3 atm K-1 mol-1
Kf of water = 1.86 K kg mol–1
Using formula (i),
m =
Using formula (ii),
π = MRT
M =
Mass of solvent =
Mass of solution = 1.063 g mL-1 × 1000 mL = 1063 g
Mass of solute = 1063 g – 897 g = 166 g
Now, using formula (iii),
m =
∴
= 334 g mol-1
The molar mass of the compound is 334 g mol-1.