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 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.

Answer 2

Answer: D. $Na_2SO_4$

Explanation:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b$ = change in boiling point

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

$K_b$ = boiling point constant

m= molality =$\frac{\text{moles of solute}}{\text{weight of solvent in kg}}=\frac{0.5}{2kg}=0.25m$ =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.