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Question
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?
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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. `triangle "T"_"f" = "K"_"f" "m"`
2. π = MRT
3. m = `(1000 "W"_2)/("M"_2 "W"_1)`
Calculation: R = 0.08205 dm3 atm K-1 mol-1
`triangle "T"_"f" = "T"_"f"^0 - "T"_"f"` = 0 °C - (- 1.03 °C) = 1.03 °C = 1.03 K
Kf of water = 1.86 K kg mol–1
Using formula (i),
`triangle "T"_"f" = "K"_"f" "m"`
m = `(triangle "T"_"f")/"K"_"f" = "1.03 K"/(1.86 "K kg mol"^-1)` = 0.554 mol kg-1 = 0.554 m
Using formula (ii),
π = MRT
M = `pi/"RT" = (12.16 "atm")/(0.08205 "dm"^3 "atm K"^-1 "mol"^-1 xx 298.15 "K")` = 0.497 mol dm-3 = 0.497 M
Mass of solvent = `(0. 497 "mol dm"^-3)/(0.554 "mol kg"^-1) xx 1 "dm"^3` = 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 = `(1000 "W"_2)/("M"_2 "W"_1)`
∴ `"M"_2 = (1000 "W"_2)/("m" "W"_1) = (1000 "g kg"^-1 xx 166 "g")/(0.554 "mol kg"^-1 xx 897 "g")`
= 334 g mol-1
The molar mass of the compound is 334 g mol-1.
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