The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 14g of the salt per 200 g of water at 100∘C is 70% .If the vapour pressure of water is 760 mmHg , what will be the vapour pressure of the solution ?
Answers
We need to find the moles of Ca(NO3)2 and water
To find vapour pressure = Po-P/Po = n/n + N
We will take one mole of Ca(NO3)2
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
Therefore, total number of moles in the solution at equilibrium
= (1 -0.7) + 0.7 + 2 ×0.7 = 2.4
Number 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)
Thus, number of moles of the solute in the solution containing 7 g of salt
n = 2.4/164×7 = 0.102
No. of moles of water (N) = Weight 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
Answer:
74.26mm
Explanation:
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
∴ Ps = 760/1.0184 = 74.26 mm