Calculate the Osmotic pressure of 0.9 % aqueous solution of solute X at 25 degree Celsius Molar mass of the solute is 60 gram per mole.(R = 0.0820 lit-atm K^-1 mol^-1).

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neelam · Answer

0.9% of aqueous solution of solute means 0.9 gram of solute X present in 100 gram of solution. It means given weight of solute is 0.9 gram. Formula of osmotic pressure isPi = n/v RT. Put the value of n= 0.9/60(given weight of solute /molar mass of solute) ,v=100, R= 0.0820 and find the answer. OK

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Calculate the Osmotic pressure of 0.9 % aqueous solution of solute X at 25 degree Celsius Molar mass of the solute is 60 gram per mole.(R = 0.0820 lit-atm K^-1 mol^-1).

Calculate the Osmotic pressure of 0.9 % aqueous solution of solute X at 25 degree Celsius Molar mass of the solute is 60 gram per mole.(R = 0.0820 lit-atm K^-1 mol^-1).

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Calculate the Osmotic pressure of 0.9 % aqueous solution of solute X at 25 degree Celsius Molar mass of the solute is 60 gram per mole.(R = 0.0820 lit-atm K^-1 mol^-1).

Calculate the Osmotic pressure of 0.9 % aqueous solution of solute X at 25 degree Celsius Molar mass of the solute is 60 gram per mole.(R = 0.0820 lit-atm K^-1 mol^-1).

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