JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Solutions

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of

2.42 \times 1 0^{- 3}

bar.
The molar mass of the biopolymer is ______

\times 1 0^{4}

g mol

^{- 1}

.
(Round off to the Nearest integer)
[Use : R

= 0.083

L bar mol

^{- 1}

^{- 1}

]

Answer: 15

Solution

For a dilute solution the osmotic pressure obeys

π V = \frac{w}{M} R T

so the molar mass is

M = \frac{w R T}{π V} = \frac{1.46 \times 0.083 \times 300}{2.42 \times 1 0 ^{- 3} \times 0.1}

M = \frac{36.354}{2.42 \times 1 0 ^{- 4}} = 1.502 \times 1 0^{5} g mo l^{- 1}

The molar mass of the biopolymer is

15 \times 1 0^{4}

mo l^{- 1}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer

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