Question

Part D the boiling point of the solution: 110 g of sucrose, C12H22011 , a nonelectrolyte, dissolved in 1.40 kg of water (Kb-o52°C) Express your answer using one decimal place. the boiling point Submit Request Answer

Answer 1

--> moles of sucrose = 110/342 = 0.32164 mol

Molality of sucrose = moles/ mass of solvent(kg)

= 0.32164/1.40 = 0.2297 m

dT = Kb*m

= 0.52*0.2297 = 0.11944 C

Boiling point of solution= 100 + 0.11944 = 100.11944 °C = 100.1°C

--> molality of glucose = (45/180)/0.130 = 1.923 m

dT = Kf*m

= 1.86*1.923 = 3.5768 C

Freezing point of solution= 0 - 3.5768 = -3.5768 °C = -3.6 °C

--> dT = Kb*m

molality of glucose= 1.923( as calculated above)

dT = 0.52*1.923 = 0.99996 T

Boiling point of solution= 100+0.99996 = 100.99996°C = 101.0°C

--> molality of sucrose = 0.2297 m(calculated above)

dT = Kf*m

= 1.86*0.2297 = 0.42724 C

Freezing point of solution= 0-0.42724 = -0.42724°C = -0.4 °C

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