CBSE Class 12 · 2011 · All India · SolutionsShort Answer

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A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

Answer: M ≈ 1.42 × 10⁴ g mol⁻¹

Solution

Convert the pressure:

π = \frac{0.335}{760} = 4.408 \times 1 0^{- 4} atm

Concentration from

π = C R T

C = \frac{4.408 \times 1 0 ^{- 4}}{0.0821 \times 298} = 1.802 \times 1 0^{- 5} mol L^{- 1}

Moles in 35.0 mL:

n = 1.802 \times 1 0^{- 5} \times 0.0350 = 6.31 \times 1 0^{- 7} mol

Molar mass:

M = \frac{8.95 \times 1 0 ^{- 3}}{6.31 \times 1 0 ^{- 7}}

M = 1.42 \times 1 0^{4} g mol^{- 1}

CBSE Class 12 Board · 2011 · All India · Subjective

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