5.Ans :-
From Raoult's law fro non-volatile solute :
Ps = P0A.XA ......................(1)
Where,
Ps = Vapor pressure of solution of glucose
P0A = Vapor pressure of pure solvent i.e. water at 100 0C is = 658 torr
XA = Mole fraction of solvent = No. of moles of solvent (nA) / Total no. of moles of solution (nA+nB)
Given,
Molarity of C6H12O6 = 3.0 M, which means 3.0 moles of C6H12O6 (nB) present in 1 L or 1000 mL of the solution.
So, Volume of the solution = 1000 mL
Given Density of the solution = 1.12 g/mL
Therefore,
Mass of the solution = Volume of the solution x Given Density of the solution = 1000 mL x 1.12 g/mL = 1120 g
Also, Mass of C6H12O6 = Moles of C6H12O6 x Gram molar mass of C6H12O6 = 3.0 mol x 180.16 g/mol = 540.48 g
So, Mass of solvent i.e. water = Mass of solution - Mass of solute
= 1120 g - 540.48 g
= 579.52 g
Therefore, No. of moles of water (nA) = Mass of water / Gram molar mass of water
= 579.52 g / 18 g/mol
= 32.1956 mol
So, XA = nA/(nA+nB)
XA = 32.1956 mol / (32.1956 mol + 3.0 mol) = 32.1956 mol / (35.1956 mol) = 0.915
Put this value in equation (1) :
Ps = 658 torr .(0.915)
Ps = 602.07 torr
Therefore, Vapor pressure of solution of glucose = Ps = 602.07 torr |
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