CBSE Class 12 · 2010 · All India Compartment; part (b) Delhi 2009 · SolutionsLong Answer

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a) Define the following terms:
i) Mole fraction
ii) van't Hoff factor
b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?

Solution

(a)(i) Mole fraction of a component = moles of that component ÷ total moles of all components of the solution.
(ii) van't Hoff factor

i

= (observed colligative property) ÷ (calculated colligative property) = (normal molar mass) ÷ (observed molar mass); it measures dissociation (

i > 1

) or association (

i < 1

).
(b) Osmotic pressure:

π = \frac{13.3}{760} = 0.0175 atm

Concentration:

C = \frac{π}{R T} = \frac{0.0175}{0.0821 \times 298} = 7.15 \times 1 0^{- 4} mol L^{- 1}

Moles in 10.0 mL =

7.15 \times 1 0^{- 6}

mol.
Molar mass:

M = \frac{0.100}{7.15 \times 1 0 ^{- 6}}

M \approx 1.40 \times 1 0^{4} g mol^{- 1}

CBSE Class 12 Board · 2010 · All India Compartment; part (b) Delhi 2009 · Subjective

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CBSE Class 12 · 2010 · All India Compartment; part (b) Delhi 2009 · SolutionsLong Answer

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