JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 16)
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 $$\times$$ 10$$-$$3 bar.
The molar mass of the biopolymer is _____________ $$\times$$ 104 g mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol$$-$$1 K$$-$$1]
The molar mass of the biopolymer is _____________ $$\times$$ 104 g mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol$$-$$1 K$$-$$1]
Answer
15
Explanation
$$\pi$$ = CRT;
$$\pi$$ = osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 $$\times$$ 10$$-$$3 bar = $${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$$
$$\Rightarrow$$ M = 15.02 $$\times$$ 104 g/mol
$$\pi$$ = osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 $$\times$$ 10$$-$$3 bar = $${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$$
$$\Rightarrow$$ M = 15.02 $$\times$$ 104 g/mol
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