We know, $$\alpha = {{i - 1} \over {n - 1}}$$

where $$\alpha$$ = degree of dissociation of an electrolyte in solution, n = no. of ions produced by 1 molecule of the electrolyte on dissociation, i = van't Hoff factor.

In aqueous solution, $$Ca{(N{O_3})_2}$$ dissociates as:

$$Ca{(N{O_3})_2}(aq) \to C{a^{2 + }}(aq) + 2NO_3^ - (aq)$$

Therefore, n = 3 and given : $$\alpha$$ = 0.7

$$\therefore$$ $$0.7 = {{i - 1} \over {3 - 1}}$$ $$\therefore$$ $$i = 2.4$$

For an electrolyte solution, the relative lowering vapour pressure is given by,

$${{{p^0} - p} \over {{p^0}}} = i \times {x_2} \approx i \times {{{n_2}} \over {{n_1}}} = 2.4 \times {{{n_2}} \over {{n_1}}}$$

In solution, no. of moles of $$Ca{(N{O_3})_2}({n_2}) = {7 \over {1.64}} = 0.042$$ mol [$$\therefore$$ Molar mass of $$Ca{(N{O_3})_2}$$ = 164 g mol^$$-$$1] and that of water $$({n_1}) = {{100} \over {18}} = 5.55$$ mol

Given : p⁰ = 760 mm Hg

$$\therefore$$ $${{760 - p} \over {760}} = {{2.4 \times 0.042} \over {5.55}}$$ $$\therefore$$ p = 746.2 mm Hg

$$\therefore$$ Vapour pressure of the solution = 746.2 mm Hg