IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

kane

the degree of dissociation of Ca(NO₃)₂ in a dil. aq. solⁿ containing 7g of salt per 100g of water at 100⁰C is 70%.calculate vapour pressure of solution?

answer is 746mm

karthik2007

Method to go about the problem:

Ca(NO₃)₂ gives 3 particles. Also. $\alpha = 0.7$ . So use $\alpha = \frac{i-1}{n-1}$ , and get the value of i.

Now, note that the vapour pressure of pure water = 760mm Hg. Hence, using the colligative property "Relative lowering of vapour pressure", we get :

$\frac{P^o - P}{P^o} = i \frac{n_{solute}}{n_{solute} + n_{solvent}}$ ,

where the solute is Ca(NO₃)₂, and the solvent is water, we get P = 746mm.

anshul_2009

mass of solute (W1)=7g

mass of solvent(W2)=100g

T=100 C = 373 K

At 373K water boils therefore

Pressure of pure solvent (Psv)=760 mm = 1atm

C=degree 0f dissociation = 70%

C= (i-1)/(a-1) i=vant hoff factor , a= no. of particles

here a=3

therefore

70/100 = (i-1)/(3-1)

0.7=(i-1)/2

1.4=i-1

i=2.4

using raoults 2nd law

(760-P)/760 = i mole fraction of solute

mole fraction of solute= (7/165)/(100/18)

kane

sorry bro but i tried the exact method but i'm not getting the answer

karthik2007

The method I've listed is correct. Of course, you could try the diluted form of the law, where u ignore the mole fraction of the solute wrt the solvent.

anshul_2009

Substitute in eqn and u wil get P=746mm , I got it try once again

kane

karthik brother & anshul i'm not doubting your method.even i tried the same method but i didn't got the answer.i think i have made some calculation mistake.

sorry guys

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