The solution containing $3 g$ of an organic compound per litre of the solution shows an osmotic pressure of $0.6 a t m$ at $127^{\circ} C$ . Calculate the molecular mass of the compound

164.2 g m o l^{- 1}

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25.7 g m o l^{- 1}

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220.1 g m o l^{- 1}

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86.4 g m o l^{- 1}

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Solution

The correct option is A $164.2 g m o l^{- 1}$
Osmotic pressure:

$π = \frac{n}{V} \times R T$

Here n is number of moles

$n = \frac{Given mass}{Molar mass} = \frac{w_{B}}{m_{B}} = \frac{3}{m_{B}}$

$V = 1 L R = 0.0821 L a t m m o l^{- 1} K^{- 1} T = 127^{\circ} C = 400 K$

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

$m_{B} = \frac{3 g \times 0.0821 L a t m m o l^{- 1} K^{- 1} \times 400 K}{0.6 a t m \times 1 L} m_{B} = 164.2 g m o l^{- 1}$

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Similar questions

2.4 a t m

127^{\circ} C

12 g

^{0}

R =

l i t r e - a t m K^{- 1} {m o l e}^{- 1}

27^{o} C

18 g

293 K

2.414 \times 10^{5} N m^{- 2}

S = 8.3 J K^{- 1}

27^{o} C

[R = 0.0822 a t m L {m o l}^{- 1}]

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