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An aqueous solution of a certain organic compound has a density of 1.063 g mL-1 , an 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? - Chemistry

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Sum

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.

Concept: Colligative Properties and Determination of Molar Mass - Osmosis and Osmotic Pressure
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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 2 Solutions
Exercises | Q 12 | Page 46
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