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An aqueous solution of glucose is prepared by dissolving solid glucose in pure water at 300 K. Through this solution, 20 L of dry nitrogen gas is passed at 1.0 atm and 300 K. As a result, the solution lost 0.45 g of weight. If density of solution is 1.24 g/cm3., determine its morality. vapor pressure of pure water at 300 K is 24 mm of Hg.

An aqueous solution of glucose is prepared by dissolving solid glucose in pure water at 300 K. Through this solution, 20 L of dry nitrogen gas is passed at 1.0 atm and 300 K. As a result, the solution lost 0.45 g of weight. If density of solution is 1.24 g/cm3., determine its morality. vapor pressure of pure water at 300 K is 24 mm of Hg.

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Grade:12

1 Answers

Arun
25750 Points
3 years ago
Here, molar mass of glucose (C6H12O6 ) = 180 gm/mol
Molar mass of water = 18 gm/mol
So, moles of sugar solute = given mass/ molar mass = 10/180 = 0.05 mol
Moles of solvent water = 90/18 = 5 mole
So, mole fraction of solvent = moles of water / moles of solution = 5/ 5+0.5 = 5/5.05 = 0.99 
Now, we have vapor pressure of solution,
Psolution = Xsolvent x P0solvent 
Where, P0solvent = pure vapor pressure of solvent (water) = 32.8mm hg (as given) 
and Xsolvent = mole fraction of solvent (water) 
So, Psolution = 0.99 x 32.8 =32.47 mmhg
 

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