Raoult’s Law-Numerical Part 2

Raoult’s Law

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Raoult’s Law-Numerical

Question 1) Cholesterol is very soluble in ether. If 0.869 gm of cholesterol is dissolved in 4.4 gm of ether (C₄ H₁₀O),the vapour pressure of ether lowers from 0.526 to 0.507 atmosphere at 18⁰ C . What is molecular weight of cholesterol ?

Solution )

p_o= 0.526 atm

P_s = 0.507 atm

p_o-p_s =0.019

w = 0.869 gm

M of  ether (C₄H₁₀O)  = 12 x 4 + 1 x 10 + 1 x 16 = 74

m of cholesterol = ?

W= 4.4 gm.

(p_o-p_s) /p_s =wM/Wm

0.019 /0.507  =0.869 x 74 / 4.4 m

m = 0.869 x 74 x 0.507 /0.019 x 4.4

m = 389.99  Ans.

Question 2) The vapour pressure of an aqueous solution of glucose at 373 K is 750 mm Hg. Calculate the molality and mole fraction of solute ?

Solution )

T =373 K 

P_o of water = 1 atmosphere = 760 mm 

P_s = 750 mm

M of water (H₂O) = 1 x 2 + 1 x 16 = 18

According to Raoult’s Law

(p_o-p_s) / p_o =wM / Wm

Molality = w x 1000/ m W

multiply by ‘M’ in numerator & denominator,

Molality = w x 1000 x M / m x W x M

because ,

(p_o-p_s ) /p_o = wM/Wm

so,

Molality =  (p_o-p_s ) x 1000 /p_o x M

                = (760 – 750) x 1000 / 760 x 18

Molality = 0.731 mole / Kg

mole fraction of solute , 

(n /n +N )= (p_o-p_s ) /p_o

                  = (760 – 750)/760 = 10 / 760 =0.013

mole fraction of solute = 0.013 Ans.

Question 3 )  Calculate the relative lowering in vapour pressure if 10 gm of a solute  ( molecular weight =100) are dissolved in 180 gm  water ?

Solution)

According to Raoult’s Law

Relative lowering in vapour pressure ,

           p_o – p_s / p_o = (n /n +N )

w =10 gm., m = 100, W = 180 gm. , M of water = 18

n = w/m = 10 / 100 =0.1

N = W/M= 180 /18 = 10

 p_o – p_s / p_o = 0.1 / (0.1 + 10)

                      = 0.1 /10.1

[ p_o – p_s / p_o ] = 0.0099

Relative lowering in vapour pressure = 0.0099   Ans.

Question 4) At 50⁰C the vapour pressure of water and ethyl alcohol are 92.5 & 219.9 mm of Hg respectively . If 6.0 gm of nonvolatile solute  (molecular weight = 120) are dissolved in 150 gm of each of these solvent. What will be the ratio of relative vapour pressure lowering in two solvents ?

Solution)

Relative lowering in vapour pressure,

           p_o – p_s / p_o = (n /n +N )

For water,

w =6.0 gm., m = 120, W = 150 gm. , M of water = 18

n = w/m = 6 / 120 =0.0.05

N = W/M= 150 /18 = 8.33

 p_o – p_s / p_o = 0.05 / (0.05+ 8.33 )

Relative lowering in vapour pressure  for water =  [ p_o – p_s / p_o] = 0.0059 

For ethyl alcohol ,

w =6.0 gm., m = 120, W = 150 gm. , M of  ethyl alcohol ( C₂H₅OH)  = 2 x 12 + 5 x 1 + 16 + 1 = 46

n = w/m = 6 / 120 =0.0.05

N = W/M= 150 /46 = 3.26

 p_o – p_s / p_o = 0.05 / (0.05+ 3.26 )

p_o – p_s / p_o = 0.015

Relative lowering in   for ethyl alcohol =  [ p_o – p_s / p_o] = 0.015 

Ratio of relative lowering vapour pressure of water & ethyl alcohol = 0.0059 / 0.015

                                                                                                                                       =0.3933 Ans.