1 . 46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2 . 42 × 10 - 3 bar The molar mass of the biopolymer is - × 10 4 g mol - 1 . (Round off to the Nearest Integer) [Use: R = 0 . 083 L bar mol - 1 K - 1

Question

1 . 46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2 . 42 × 10 - 3 bar The molar mass of the biopolymer is - × 10 4 g mol - 1 . (Round off to the Nearest Integer) [Use: R = 0 . 083 L bar mol - 1 K - 1,

MARKS App · Accepted Answer

π = CRT ; π = osmotic pressure C = molarity T = Temperature of solution let the molar mass be Molar mass of polymer = M gm / mol 2 . 42 × 10 - 3 bar = = 1 . 46 g M gm / mol 0 . 1 ℓ × 0 . 083 ℓ . bar mol . K × 300 K ⇒ M = 15 . 02 × 10 4 g / mol

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