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Standard XIII

Chemistry

Osmotic Pressure

A solution of...

Question

A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 lit. atm.k^-1mol^-1) at 273 K?

6.02 atm

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4.92 atm

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4.04 atm

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5.32 atm

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Solution

Weight given (sucrose) = 68.4 gram
Gas constant (R) = 0.082 lit. atm. $K^{- 1}$ ${m o l}^{- 1}$
Temperature = 273 K
Molecular mass = 342 gm/mol
V = 1
Hence using the formula of osmotic pressure we get
$π = \frac{w \times R \times T}{m o l a r m a s s \times V o l u m e}$
$π = \frac{68.4 \times 0.082 \times 273}{342}$
$π = 4.48 a t m$

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Q. A solution of sucrose (Molar mass

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A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 lit. atm.k^-1mol^-1) at 273 K?

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A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 lit. atm.k-1mol-1) at 273 K?

Weight given (sucrose) = 68.4 gramGas constant (R) = 0.082 lit. atm.K−1 mol−1 Temperature = 273 KMolecular mass = 342 gm/molV = 1Hence using the formula of osmotic pressure we getπ=w×R×Tmolarmass×Volumeπ=68.4×0.082×273342 π=4.48atm

A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 lit. atm.k^-1mol^-1) at 273 K?