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Anton [14]
3 years ago
7

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 boil

ing point of the solution? A. NaCl B. KI C. CO2 D. Na2SO4
Chemistry
2 answers:
serious [3.7K]3 years ago
8 0

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.

arlik [135]3 years ago
7 0

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.

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Answer:

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The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

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