$500 c m^{3}$ of an aqueous solution contains $0.5 g$ of a polymer. The osmotic pressure of such solution at $300 K$ is found to be $2 \times 10^{- 3} a t m$ . What will be the molar mass of the polymer ?

25.46 k g m o l^{- 1}

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12.32 k g m o l^{- 1}

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123.33 k g m o l^{- 1}

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0.23 k g m o l^{- 1}

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Solution

The correct option is B $12.32 k g m o l^{- 1}$
Osmotic pressure:

$π = \frac{n}{V} \times R T$

Here n is number of moles

$n = \frac{Given mass}{Molar mass} = \frac{w_{B}}{m_{B}}$

$V = 500 c m^{3} = 0.5 L R = 0.0821 L a t m m o l^{- 1} K^{- 1}$

$\Rightarrow π V = \frac{w_{B}}{m_{B}} \times R T m_{B} = \frac{w_{B} \times R \times T}{π \times V}$

$m_{B} = \frac{0.5 g \times 0.0821 L a t m m o l^{- 1} K^{- 1} \times 300 K}{2 \times 10^{- 3} a t m \times 0.5 L} m_{B} = 12315 g m o l^{- 1} m_{B} = 12.315 k g m o l^{- 1}$

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300 c m^{3}

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2.57 \times 10^{- 3}

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2.57 \times 10^{- 3} b a r

^{- 1}

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2.57 \times 10^{- 3}

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