The osmotic pressure of a polymer in a solvent at 300 K was measured at two concentrations: pi = 1.385 x 10^-3 atm at 2.0 g L^-1 and pi = 6.556 x 10^-3 atm at 6.0 g L^-1. For real polymer solutions pi/c varies linearly with c (in g L^-1), and the intercept at c -> 0 equals RT/M. Using R = 0.0821 L atm mol^-1 K^-1, find the molar mass of the polymer in g mol^-1 (nearest thousand).

JEE Advanced levelNumerical answerDifficulty 3 of 3so089

The osmotic pressure of a polymer in a solvent at $300 K$ was measured at two concentrations: $π = 1.385 \times 1 0^{- 3} atm$ at $2.0 g L^{- 1}$ and $π = 6.556 \times 1 0^{- 3} atm$ at $6.0 g L^{- 1}$ . For real polymer solutions $π / c$ varies linearly with $c$ (in $g L^{- 1}$ ), and the intercept at $c \to 0$ equals $RT / M$ . Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , find the molar mass of the polymer in $g mo l^{- 1}$ (nearest thousand).

Numerical answer type, answer in g/mol, tolerance 600.

π / c

6.925 \times 1 0^{- 4}

2.0

and

10.927 \times 1 0^{- 4}

6.0 g L^{- 1}

(in

atm L g^{- 1}

). Slope

= (10.927 - 6.925) \times 1 0^{- 4} /4.0 = 1.000 \times 1 0^{- 4}

, so the intercept is

4.924 \times 1 0^{- 4}

. Then

M = RT / intercept = 0.0821 \times 300/ (4.924 \times 1 0^{- 4}) \approx 50019 g mo l^{- 1}