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 x - 104 g mol-1 Round off to the Nearest Integer -Use - R-0-083 L bar mol-1K-1-

Osmotic pressure: (π) π=i× C × R× T Here i is van't Hoff factor nbsp; C is concentration (Molarity) R is ideal gas constant nbsp; nbsp; T is temperatur ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

1.46 g of a b...

Question

$1.46 g$ of a biopolymer dissolved in a $100 m L$ water at $300 K$ exerted an osmotic pressure of $2.42 \times 10^{- 3}$ bar.

The molar mass of the biopolymer is $x \times 10^{4} g {mol}^{- 1}$ (Round off to the Nearest Integer)

[Use : $R = 0.083 L {bar mol}^{- 1} K^{- 1}$ ]

15.00

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Solution

Osmotic pressure: $(π)$
$π = i \times C \times R \times T$
Here
$i is van't Hoff factor C is concentration (Molarity) R is ideal gas constant T is temperature$

$2.42 \times 10^{- 3} = \frac{1 \times 1.46 \times 1000 \times 0.083 \times 300}{M \times 100}$

$M = 150223 g {mol}^{- 1}$

$M = 15.0223 \times 10^{4} {g mol}^{- 1} x = 15$

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1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42×10−3 bar. The molar mass of the biopolymer is x×104 g mol−1 (Round off to the Nearest Integer) [Use : R=0.083 L bar mol−1K−1]

Osmotic pressure: (π) π=i×C×R×T Here i is van't Hoff factorC is concentration (Molarity)R is ideal gas constant T is temperature 2.42×10−3=1×1.46×1000×0.083×300M×100 M=150223 g mol−1 M=15.0223×104 g mol−1x=15

$1.46 g$ of a biopolymer dissolved in a $100 m L$ water at $300 K$ exerted an osmotic pressure of $2.42 \times 10^{- 3}$ bar.

The molar mass of the biopolymer is $x \times 10^{4} g {mol}^{- 1}$ (Round off to the Nearest Integer)

[Use : $R = 0.083 L {bar mol}^{- 1} K^{- 1}$ ]