Question #3c472

1 Answer
Aug 21, 2016

Answer:

Heres what I get.

Explanation:

(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#