First find moles of Ca(NO3)2 and water. Then use the expression
Po-P/Po = n/n + N to find vapour pressure of solution
Let initially 1 mole of Ca(NO3)2 is taken
Degree of dissociation of Ca(NO3)2 = 70/100 = 0.7
Ionization of Ca(NO3)2 can be represent as
Ca(NO3)2 ⇌ Ca2+ + 2NO-3
At start 1 0 0
At equilibrium 1 -0.7 0.7 2 * 0.7
∴ Total number of moles in the solution at equilibrium
= (1 -0.7) + 0.7 + 2 * 0.7 = 2.4
No. of moles when the solution contains 1 gm of calcium nitrate instead of 1 mole of the salt
= 2.4/164 (164 is the mol. wt. of Cal. nitrate)
∴ No. of moles of the solute in the solution containing 7 g of salt, i.e.,
n = 2.4/164 * 7 = 0.102
No. of moles of water (N) = Wt. of water/Mol. wt. of water = 100/18 = 5.55
Applying Raoult’s law, Po-P/Po = n/n + N
760 –p/760 = 0.102/0.102 + 5.55
760 – p/760 = 0.0180
⇒ p = 760 – (760 * 0.0180) = 746.3 mm Hg
ALTERNATIVESOLUTION :
Ca(NO3)2 ⇌ Ca2+ + 2NO-3
1 0 0 before dissociation
1 - ∝ ∝ 2∝ After dissociation
∴ Total moles at equilibrium = (1 +2∝)
= 1 + 2 * 0.7 (∵ a = 0.7)
= 2.4
For Ca(NO3)2 : mob/mexp = 1 + 2∝
∴ mexp = mob/1 + 2 *0.7 = 164/2.4 = 68.33
Also at 100° PoH base 2O = 760 mm, w = 7g
W = 100 g
Now, Po–Ps/Ps = 7 *18/68.33 *100 = 0.0184
Or Po/Ps – 1 = 0.0184
∴ Ps = 760/1.0184 = 74.26 mm
