Part A 0.117m K2S Submit My Answers Give Up Part B 21.6g of CuCl2 in 465g water

Question

Part A

0.117m K2S

SubmitMy AnswersGive Up

Part B

21.6g of CuCl2 in 465g water

SubmitMy AnswersGive Up

Part C

6.3% NaNO3 by mass (in water)

Express your answer using two significant figures.

SubmitMy AnswersGive Up

Part D

Calculate the boiling point of the solution in part A, assuming complete dissociation.

Express your answer using six significant figures.

SubmitMy AnswersGive Up

Part E

Calculate the boiling point of the solution in part B, assuming complete dissociation.

Express your answer using six significant figures.

SubmitMy AnswersGive Up

Part F

Calculate the boiling point of the solution in part C, assuming complete dissociation.

Express your answer using five significant figures.

SubmitMy AnswersGive Up

Explanation / Answer

(A) K2S => 2 K+ + S2-

van't Hoff factor i = 3 (1 K2S gives 3 ions)

Freezing point depression DTf = i x Kf x molality

= 3 x 1.86 x 0.117 = 0.653 deg C

Freezing point of solution Tf = freezing point of water - DTf

= 0.00 - 0.653 = -0.653 deg C

(B) CuCl2 => Cu2+ + 2 Cl-

van't Hoff factor i = 3 (1 CuCl2 gives 3 ions)

Moles of CuCl2 = mass/molar mass of CuCl2

= 21.6/134.45 = 0.16065 mol

Molality = moles of CuCl2/mass of water in kg

= 0.16065/0.465 = 0.3455 m

Freezing point depression DTf = i x Kf x molality

= 3 x 1.86 x 0.3455 = 1.93 deg C

Freezing point of solution Tf = freezing point of water - DTf

= 0.00 - 1.93 = -1.93 deg C

(C) NaNO3 => Na+ + NO3-

van't Hoff factor i = 2 (1 NaNO3 gives 2 ions)

Consider 100 g of solution

Moles of NaNO3 = mass/molar mass of NaNO3

= (6.3/100 x 100)/84.995 = 0.074122 mol

Mass of water = 100 - 6.3 = 93.7 g = 0.0937 kg

Molality = moles of NaNO3/mass of water in kg

= 0.074122/0.0937 = 0.79106 m

Freezing point depression DTf = i x Kf x molality

= 2 x 1.86 x 0.79106 = 2.94 deg C

Freezing point of solution Tf = freezing point of water - DTf

= 0.00 - 2.94 = -2.94 deg C

(D) Boiling point elevation DTb = i x Kb x molality

= 3 x 0.512 x 0.117 = 0.180 deg C

Boiling point of solution Tb = boiling point of water + DTb

= 100 + 0.180 = 100.180 deg C

(E) Boiling point elevation DTb = i x Kb x molality

= 3 x 0.512 x 0.3455 = 0.531 deg C

Boiling point of solution Tb = boiling point of water + DTb

= 100 + 0.531 = 100.531 deg C

(F) Boiling point elevation DTb = i x Kb x molality

= 2 x 0.512 x 0.79106 = 0.810 deg C

Boiling point of solution Tb = boiling point of water + DTb

= 100 + 0.810 = 100.810 deg C

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.