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natka813 [3]
2 years ago
8

a 70 g sample of cesium is sealed in a glass vial and lowered into 250 mL of water at 90 C. when the cesium melted the temperatu

re of the water dropped to 88.98 C
Chemistry
1 answer:
andre [41]2 years ago
7 0

Answer: 70.0 g  = .53 mol  

Q = 250g (+ 1.02) 4.18

Q = 1065. 9 J / 1000 = 1.0659 kJ

1.0659/ .53 = 2.01 kJ/ mol

Answer: 2.01 kJ/ mol

Explanation:

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3 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

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Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

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