CBSE Class 12 · 2006 · All India Compartment · SolutionsShort Answer

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(a) A 4% solution of sucrose is isotonic with a 3% solution of an unknown organic substance. Calculate the molecular mass of the unknown substance.
(b) What is the mole fraction of the solute in a 2.5 m aqueous solution?

Answer: (a) M ≈ 256.5 g mol⁻¹ (b) x₂ ≈ 0.043

Solution

(a) Isotonic solutions have equal osmotic pressures, hence equal molar concentrations:

\frac{4}{342} = \frac{3}{M}

M = \frac{3 \times 342}{4}

M = 256.5 g mol^{- 1}

(b) 2.5 m: 2.5 mol solute per 1000 g water (

= 1000/18 = 55.56

mol):

x_{2} = \frac{2.5}{2.5 + 55.56} = \frac{2.5}{58.06}

x_{2} = 0.043

CBSE Class 12 Board · 2006 · All India Compartment · Subjective

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