A solution containing 3.00 g of calcium nitrate in 100 c.c. of solution had an osmotic pressure of 11.2 atmosphere at 12 C. Calculate the degree of ionisation of calcium nitrate at this dilution and temperature??
Answers
Please find the answer to your question
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