Solutions

Solutions

Question

(a) State Raoult’s law for a solution containing volatile components.

How does Raoult’s law become a special case of Henry’s law?

(b) 1·00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0·40 K. Find the molar mass of the solute. (Kf for benzene = 5·12 K kg mol-1

Answer

(a) Let p1p2 = Partial vapour pressure of two volatile components 1 and 2 of a mixture straight p subscript 1 superscript 0 space straight p subscript 2 superscript 0

= Vapour pressure of pure components 1 and 2

x1x= Mole fractions of the components 1 and 2

ptotal = Total vapour pressure of the mixture then Raoult’s law can be stated as: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

That is, for component 1,

p1  straight alpha    x1

And, p1 = straight p subscript 1 superscript 0x1

For component 2,

P2 = straight p subscript 2 superscript 0x2

According to Dalton’s law of partial pressures,

straight p subscript total space equals space straight p subscript 1 space plus straight p subscript 2
space space space space space space equals space straight p subscript 1 superscript 0 space straight x subscript 1 space plus straight p subscript 2 superscript 0 straight x subscript 2
space space space space space space equals space left parenthesis 1 minus straight x subscript 2 right parenthesis straight p subscript 1 superscript 0 space plus straight p subscript 2 superscript 0 space straight x subscript 2
straight P subscript total space equals space straight p subscript 1 superscript 0 space plus left parenthesis straight p subscript 2 superscript 0 space minus straight p subscript 1 superscript 0 right parenthesis space straight x subscript 2

 The plot of vapour pressure and mole fraction of an ideal solution at constant temperature is shown below.

 

 

Raoult’s Law as a Special Case of Henry’s Law

According to Raoult’s law, the vapour pressure of a volatile component in a given solution is p1 =straight p subscript 1 superscript 0 x1

According to Henry’s law, the partial vapour pressure of a gas (the component is so volatile that it exists as gas) in a liquid is

p = KH x

It can be observed that in both the equations, the partial vapour pressure of the volatile component varies directly with its mole fraction. Only the proportionality constants KH and are different. Thus, Raoult’s law becomes a special case of Henry’s law in which KH is equal to .

(b) Given: w2 = 1g (weight of solute)

w1 = 50 g (weight of solvent)

Tf = 0.40 K

kf = 5.12 K Kg mol-1

M2 =? (Molar mass of solute)

Using the equation,

Tf = Kfm (where m is molality)

0.40 = 5.12 x m

 straight m space equals fraction numerator 0.40 over denominator 5.12 end fraction

straight m space equals 0.078 space mol space kg to the power of negative 1 end exponent

straight m space equals fraction numerator straight w subscript 2 divided by straight M subscript 2 over denominator straight w subscript 1 divided by 1000 end fraction

0.078 space equals fraction numerator 1 space straight x space 1000 over denominator straight M subscript 2 space straight x space 50 end fraction

straight M subscript 2 space equals fraction numerator 1000 over denominator 50 space straight x space 0.078 end fraction

straight M subscript 2 space equals space 256.41 space Kg space mol to the power of negative 1 end exponent

 

 

 

 

 

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