Raoult’s law- Numerical part 3

Raoult’s law-

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Raoult’s law-

Question 1) At 300 K , the vapour pressure of an ideal solution containing one mole of A and 3 moles of B is 550 mm. At the same temperature if  one mole of B is added to the solution , the vapour pressure of solution increases by 10 mm Hg. Calculate the vapour pressure of A and B in their respective pure state ?

Solution)

Initially,

P_M =550 mm, P_A⁰ = ? , P_B⁰ =?, n_A = 1 , n_B = 3

P_M = P_A⁰.X_A + P_B⁰.X_B

= P_A⁰ [n_A/(n_A + n_B)] +  P_B⁰ [n_B/(n_A + n_B)]

550 = P_A⁰ [1/(1 + 3)] +  P_B⁰ [3 /(1 + 3)]

550 = ( P_A⁰/4 ) + ( 3P_B⁰/ 4)

550 x 4 = P_A⁰ + 3P_B⁰

P_A⁰ + 3P_B⁰ =2200        ——– eq.1

On addition of one mole of B,

n_A = 1 , n_B = 3+1 = 4

The vapour pressure of solution increases by 10 mm Hg . Thus ,

P_M =550 + 10 = 560 mm

P_M = P_A⁰.X_A + P_B⁰.X_B

= P_A⁰ [n_A/(n_A + n_B)] +  P_B⁰ [n_B/(n_A + n_B)]

560 = P_A⁰ [1/(1 + 4)] +  P_B⁰ [4 /(1 + 4)]

560 = ( P_A⁰/5 ) + ( 4 P_B⁰/ 5)

560 x 5 = P_A⁰ + 4P_B⁰

P_A⁰ + 4 P_B⁰ =2800        ——– eq. 2

P_A⁰ + 3 P_B⁰ =2200

P_A⁰ + 4 P_B⁰ =2800

By Subtracting ,

– P_B⁰ =-600

 P_B⁰ = 600 mm Ans.

putting the value of P_B⁰ in eq. 1

P_A⁰ + 3 x 600 = 2200

P_A⁰ = 2200 – 1800 = 400 mm

P_A⁰ = 400 mm Ans.

Question 2) 0.1 M solution of glucose was found to be isotonic with a solution of X in 100 gm  water. Calculate relative lowering in vapour pressure of solution of X in water ?

Solution )

0.1 M solution of glucose is isotonic with solution of X in 100 gm water.

Therefore , concentration of solution of X = 0.1 M

It means 0.1 mole of X in 1000 ml or one litre of solution.

For dilute solution ,

volume of solution = volume of solvent (water) = weight of water( = v x d = 1000 x 1= 1000 gm)

W = 1000 gm , M of water = 18

mole of X  ‘n’= 0.1, mole of water  ‘N’ =W/M = 1000/18

According to Raoult’s law

 p_o – p_s / p_o = (n /n +N )

= 0.1 / [ 0.1 + (1000/18)]

Relative lowering of vapour pressure = p_o – p_s / p_o = 0.0018 Ans.

Question 3) What weight of solute  ( molecular weight =60) is required to dissolve in 180 gm of water to reduce the vapour pressure to 4/5^th of pure water ?

Solution )

p_s = 4/5 p_o , w = ? , m = 60 , W = 180 gm , M of water = 18

According to Raoult’s law

(p_o-p_s) /p_o =wM/Wm

[p_o– (4p_o /5) /p_o] = w x 18 /180 x 60

(5 p_o – 4 p_o) /5 p_o = w /10 x 60

1 /5 = w / 600

w = 600 /5 = 120

w = 120 gm.

Question 4 )   A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 27 ⁰C.When mole fraction of ethyl alcohol is 0.65. Calculate the vapour pressure of ethyl alcohol, if vapour pressure  of propyl alcohol is 210 mm.

Solution )

P_M = 290 mm , P^o_C2H5OH= ? ,

P_M = P^o_C2H5OH.X^o_C2H5OH+ P^o_C3H7OH.X^o_C3H7OH

X _C2H5OH = 0.65 , X_C3H7OH = 1 –  0.65 = 0.35 , P^o_C3H7OH = 210 mm

290  = P^o_C2H5OH x 0.65 + 0.35 x 210

290  = P^o_C2H5OH x 0.65 + 73.5

290 – 73.5 = P^o_C2H5OH x 0.65

P^o_C2H5OH  = 216.5 / 0.65 = 333.07

P^o_C2H5OH  = 333.o7 Ans.