The degree of dissociation of Ca(NO3 )2 in dilute solution aqueous solution containing 7.0g of the solute per 100g of water at 1000 C is 70 percent. If the vapour pressure of water at 1000 C is 760mm, calculate the vapour pressure of the solution.

Molecular mass of Ca(NO3)2 = 164g/molTherefore moles of Ca(NO3)2 = 7/164 = 0.042 moles Ca(NO3)2 Ca2+ 2NO32- 1 mol 1 mol 2 molMoles dissolved m 0 0After dissociation m(1-a) ma 2maTotal moles present = m(1-a) + ma + 2ma = m (1+2a)= 0.042 (1 + 2(0.7))= 0.102 molesMass of water = 100-7 = 93gMoles of water = 93/18 = 5.167 molesTotal moles present in solution = 0.102 + 5.167 = 5.269 molesMole fraction of water in solution = Moles of water / Total moles = 5.167/5.269pa = p°a - xa= 760 * 5.167/5.269= 745.3 mm Hg