(i) #"1 mol/L NaOH"# ≈ #"1 mol/L NH"_4"NO"_3# ≈ #"1 mol/L KNO"_3# < #"1 mol/L Na"_2"SO"_4#
(ii) #"0.2 mol/L Na"_2"CO"_3# < #"0.1 mol/L" ("NH"_4)_2"SO"_4."FeSO"_4."6H"_2"O"# < #"0.2 mol/L NaOH"# < #"0.1 mol/L AgNO"_3#
(i) Boiling points
The formula for calculating boiling point elevation is
#color(blue)(|bar(ul(color(white)(a/a) ΔT_"b" = iK_"b"m color(white)(a/a)|)))" "#
where
#ΔT_"b"# is the increase in freezing point
#i# is the van't Hoff #i# factor
#K_"b"# is the molal boiling point elevation constant
#m# is the molality of the solution.
In each of these solutions, #K_"b"# is a constant, and the molality is almost the same as the molarity.
∴ #ΔT_"b" ∝ iM#
#"For 1 mol/L NaOH", iM = 2 × 1 = 2#
#"For 1 mol/L Na"_2"SO"_4, iM = 3 × 1 = 3#
#"For 1 mol/L NH"_4"NO"_3, iM = 2 × 1 = 2#
#"For 1 mol/L KNO"_3, iM = 2 × 1 = 2#
#ΔT_"b"# is greatest for #"Na"_2"SO"_4#, so #"Na"_2"SO"_4# has the highest boiling point.
The order of boiling points is
#"1 mol/L NaOH"# ≈ #"1 mol/L NH"_4"NO"_3# ≈ #"1 mol/L KNO"_3# < #"1 mol/L Na"_2"SO"_4#
(ii) Freezing points
The formula for calculating freezing point depression is
#color(blue)(|bar(ul(color(white)(a/a) ΔT_"f" = iK_"f"m color(white)(a/a)|)))" "#
where
#ΔT_"f"# is the decrease in freezing point and
#K_"f"# is the molal freezing point depression constant
By the same argument as before
∴ #ΔT_"f" ∝ iM#
#"For 0.2 mol/L NaOH", iM = 2 × 0.2 = 0.4#
#"For 0.2 mol/L Na"_2"CO"_3, iM = 3 × 0.2 = 0.6#
#"For 0.1 mol/L AgNO"_3, iM = 2 × 0.1 = 0.2#
#"For 0.1 mol/L" ("NH"_4)_2"SO"_4."FeSO"_4."6H"_2"O", iM = 5 × 0.1 = 0.5#
#ΔT_"f"# is greatest for #"Na"_2"CO"_3#, so the #"Na"_2"CO"_3# has the lowest freezing point.
The order of freezing points is
#"0.2 mol/L Na"_2"CO"_3# < #"0.1 mol/L" ("NH"_4)_2"SO"_4."FeSO"_4."6H"_2"O"# < #"0.2 mol/L NaOH"# < #"0.1 mol/L AgNO"_3#
