degree of dissociation of Ca(NO)3 in dilute aqueous solution containing 7g of solute per 100g of water at 100degree celsius is 70%. if the vapour pressure of water at 100degree celsius is 760mm of mercury, calculate the vapour pressure of the solution

Molecular mass of Ca(NO3)2 = 164g/mol

Therefore moles of Ca(NO3)2 = 7/164 = 0.042 moles

          Ca(NO3)2         Ca2+         2NO32-

          1 mol            1 mol          2 mol

Moles dissolved  m              0             0

After dissociation m(1-a)          ma             2ma

Total moles present = m(1-a) + ma + 2ma = m (1+2a)

= 0.042 (1 + 2(0.7))

= 0.102 moles

Mass of water = 100-7 = 93g

Moles of water = 93/18 = 5.167 moles

Total moles present in solution = 0.102 + 5.167 = 5.269 moles

Mole fraction of water in solution = Moles of water / Total moles = 5.167/5.269

p_a = p°a - x_a

= 760 * 5.167/5.269

= 745.3 mm Hg