Question

An aqueous solution of glucose is prepared by dissolving solid glucose in pure water at 300K. Through this solution, 20L of dry Nitrogen gas is passed at 1.0 atm and 300K. As a result, the solution lost 0.45 g of weight. If density of solution is 1.24g/cm^3, determine its molarity. Vapour pressure of pure water at 300K is 24mm of Hg

Answer 1

Explanation:

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