11. A solution is prepared by mixing 50.0 g glucose (C6H1206) with 600.0 g of wa

Question

11. A solution is prepared by mixing 50.0 g glucose (C6H1206) with 600.0 g of water. What is the vapor pressure of this solution at 25 degree C? [At 25 degree C the vapor pressure of pure water is 23.8 torr. Glucose is a nonvolatile, nonelectrolyte.] 12. Calculate the boiling point elevation of a 0.115 m Na2SO4 aqueous solution. [Kb (H2O) = 0.51 degree C kg/mol] 13. What mass of glycerin (C3H803), a nonelectrolyte, must be dissolved in 200.0 g water to give a solution with a freezing point of-1.50 degree C?[kr(H2O) = 1.86 degree C kg/mo I]

Explanation / Answer

the formula to calculate boiling point elevation is

elevation in boiling point (dTb) = i x Kb x m

so

dTb = i x Kb x m

here

i = vanthoffs factor

Kb = a constant

m = molality

Now given solute is Na2S04

the dissociation of Na2S04 is given by

Na2S04 ----> 2Na+ + S042-

now

vanthoff factor represents the number of particles after dissociation

in the above reaction

after dissociation

there are two Na+ ions and one S042- ion

total of three particles

so

the value of i is 3

i= 3

now

dTb = 3 x 0.51 x 0.115

dTb = 0.17595

dTb = 0.18 C

so

the boiling point elevation of 0.115m Na2S04 solution is 0.18 C

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