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  • 1.46 g of a biopolymer dissolved in a 100mL water at 300 K  exerted an osmotic pressure of The molar mass of the bio

1.46 g of a biopolymer dissolved in a 100mL water at 300 K  exerted an osmotic pressure of 2.42 \times 10^{-3} \mathrm{bar}.
The molar mass of the biopolymer is_____\mathrm{\times 10^4} \mathrm{mol}^{-1}. (Round off to the Nearest Integer) .
[Use : \mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}]
 

Answers (1)

best_answer

As we have learnt,

\mathrm{\mathrm{\pi}=C R T}

Now,

\begin{aligned} &\mathrm{C=\frac{1.46}{M \times 0.1}=\frac{14.6}{M}} \\ &\pi=2.42 \times 10^{-3} \text { bar } \\ &\mathrm{R=0.083 \text { L~Bar } \mathrm{mol}^{-1} \mathrm{~K}^{-1}} \\ &\mathrm{T=300 \mathrm{~K} }\end{aligned}

Putting these values in the formula, we have

\begin{aligned} &\mathrm{2.42 \times 10^{-3}=\frac{14.6}{M} \times 0.083 \times 300} \\ & \mathrm{M=\frac{363.54}{2.42 \times 10^{-3}}=150223.14} \\ &\mathrm{M=15.02 \times 10^{4} \mathrm{~g}} \end{aligned}

Hence, the correct answer is 15.

Posted by

sudhir.kumar

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