Solutions

Solutions

Question

(a) Explain the following:

(i) Henry’s law about the dissolution of a gas in a liquid.

(ii) Boiling point elevation constant for a solvent.

(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass if glycerol was dissolved to make this solution? (Kb for water = 0.512 K kg mol-1)

Answer

(a)

(i) Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:

p = KHx

Where,

KH is Henry’s law constant

(ii) The boiling point elevation constant or molal elevation constant is a constant quantity for a solute which is related to molar mass and elevation in boiling point by the following relation.

 straight K subscript straight b straight space equals straight space fraction numerator increment straight T subscript straight b straight space cross times straight space straight M subscript straight B straight space cross times straight space straight W subscript straight A over denominator straight W subscript straight B straight space cross times straight space 1000 end fraction

Where, Kb is the boiling point elevation constant

MB is the molar mass of the solute

WB is the weight of the solute

WA is the weight of the solvent

Tb is the elevation in boiling point

(b) WB =?

WA = 500g

Kb = 0.512 Kkg mol-1

Tb = 100.42°C - 100°C

= 0.42°C

MB = 3 x 12 + 8 x 1 + 3 x 16

= 36 + 8 + 48 = 92

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