(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
