Question 3c472

Aug 21, 2016

Heres what I get.

Explanation:

(i) $\text{1 mol/L NaOH}$${\text{1 mol/L NH"_4"NO}}_{3}$${\text{1 mol/L KNO}}_{3}$ < ${\text{1 mol/L Na"_2"SO}}_{4}$

(ii) ${\text{0.2 mol/L Na"_2"CO}}_{3}$ < $\text{0.1 mol/L" ("NH"_4)_2"SO"_4."FeSO"_4."6H"_2"O}$ < $\text{0.2 mol/L NaOH}$ < ${\text{0.1 mol/L AgNO}}_{3}$

(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
${K}_{\text{b}}$ is the molal boiling point elevation constant
$m$ is the molality of the solution.

In each of these solutions, ${K}_{\text{b}}$ 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 ${\text{Na"_2"SO}}_{4}$, so ${\text{Na"_2"SO}}_{4}$ has the highest boiling point.

The order of boiling points is

$\text{1 mol/L NaOH}$${\text{1 mol/L NH"_4"NO}}_{3}$${\text{1 mol/L KNO}}_{3}$ < ${\text{1 mol/L Na"_2"SO}}_{4}$

(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
${K}_{\text{f}}$ 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 ${\text{Na"_2"CO}}_{3}$, so the ${\text{Na"_2"CO}}_{3}$ has the lowest freezing point.

The order of freezing points is

${\text{0.2 mol/L Na"_2"CO}}_{3}$ < $\text{0.1 mol/L" ("NH"_4)_2"SO"_4."FeSO"_4."6H"_2"O}$ < $\text{0.2 mol/L NaOH}$ < ${\text{0.1 mol/L AgNO}}_{3}$