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(A) The relative decrease in vapor pressure is given by Raoult's law: $\frac{\Delta P}{P^o} = X_{	ext{solute}} = \frac{i \cdot n_{	ext{solute}}}{i \

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$iv < i < ii < iii$

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(A) The relative decrease in vapor pressure is given by Raoult's law: $\frac{\Delta P}{P^o} = X_{	ext{solute}} = \frac{i \cdot n_{	ext{solute}}}{i \cdot n_{	ext{solute}} + n_{	ext{solvent}}}$.Since the moles of solute and solvent are constant,the relative decrease is directly proportional to the van't Hoff factor $(i)$.For $(i) \ NaCl$,$i = 2$ $(Na^+ + Cl^-)$.For $(ii) \ K_2SO_4$,$i = 3$ $(2K^+ + SO_4^{2-})$.For $(iii) \ Na_3PO_4$,$i = 4$ $(3Na^+ + PO_4^{3-})$.For $(iv) \ 	ext{glucose}$,$i = 1$ (non-electrolyte).Thus,the order of $i$ is $iv (1) Therefore,the order of relative decrease in vapor pressure is $iv < i < ii < iii$.

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

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$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:$(i) \ NaCl$$(ii) \ K_2SO_4$$(iii) \ Na_3PO_4$$(iv) \ \text{glucose}$Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order: