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_{A}^{0} = ? , P_{B}^{0} =?, n_{A} = 1 , n_{B} = 3*

*P _{M} = P_{A}^{0}.X_{A} + P_{B}^{0}.X_{B}*

*= P _{A}^{0} [n_{A}/(n_{A} + n_{B})] + P_{B}^{0} [n_{B}/(n_{A} + n_{B})]*

*550 = P _{A}^{0} [1/(1 + 3)] + P_{B}^{0} [3 /(1 + 3)]*

*550 = ( P _{A}^{0}/4 ) + ( 3P_{B}^{0}/ 4)*

*550 x 4 = P _{A}^{0} + 3P_{B}^{0}*

*P _{A}^{0} + 3P_{B}^{0} =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_{A}^{0}.X_{A} + P_{B}^{0}.X_{B}*

*= P _{A}^{0} [n_{A}/(n_{A} + n_{B})] + P_{B}^{0} [n_{B}/(n_{A} + n_{B})]*

*560 = P _{A}^{0} [1/(1 + 4)] + P_{B}^{0} [4 /(1 + 4)]*

*560 = ( P _{A}^{0}/5 ) + ( 4 P_{B}^{0}/ 5)*

*560 x 5 = P _{A}^{0} + 4P_{B}^{0}*

*P _{A}^{0} + 4 P_{B}^{0} =2800 ——– eq. 2*

*P _{A}^{0} + 3 P_{B}^{0} =2200*

*P _{A}^{0} + 4 P_{B}^{0} =2800*

*By Subtracting ,*

*– P _{B}^{0} =-600*

## P_{B}^{0} = 600 mm Ans.

*putting the value of P _{B}^{0} in eq. 1*

*P _{A}^{0} + 3 x 600 = 2200*

*P _{A}^{0} = 2200 – 1800 = 400 mm*

## P_{A}^{0} = 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.

_{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 ^{0}C.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*