"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]"

Question

Accepted Answer

"$\pi$ = 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"

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]

Solution

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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]

Solution

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More States of Matter questions

Drill 25 more like these. Every day. Free.

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]