Question

If 0.500 mol of each of the following solutes is dissolved in 2.0 L of water, which will cause the greatest increase in the boiling point of the solution? A. NaCl B. KI C. CO2 D. Na2SO4

Answer 1

Answer: D.
`Na_2SO_4`

Explanation:

`\Delta T_b=i* K_b* m`

`\Delta T_b` = change in boiling point

i= vant hoff factor = no of ions produced on complete dissociation

`K_b` = boiling point constant

m= molality =
`\frac{\text{moles of solute}}{\text{weight of solvent in kg}}=(0.5)/(2kg)=0.25m` =same for all solutes

1. For NaCl:

`NaCl\rightarrow Na^++Cl^-`

i=2

2. For
`KI` ,

`KI\rightarrow K^++I^-`

i=2

3.
`CO_2` , i= 1 as it does not dissociate

4. For
`Na_2SO_4` ,

`Na_2SO_4\rightarrow 2Na^++SO_4^(2-)`

i=3

Thus elevation in boiling point will be highest in
`Na_2SO_4` solution.

Answer 2

Answer is: D. Na2SO4.

b(solution) = 0.500 mol ÷ 2.0 L.

b(solution) = 0.250 mol/L.

b(solution) = 0.250 m; molality of the solutions.

ΔT = Kf · b(solution) · i.

Kf - the freezing point depression constant.

i - Van 't Hoff factor.

Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).

Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.

Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.

Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).

Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.