Question

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Answer 1

hey!!!
________

here, we have

w2 = 0.456g

M2 = 152

w1 = 31.4g

To = 56.30℃

Kb = 17.2℃/100g

∆Tb = 100Kb.w2 / w1M2

= 100×17.2×0.456 / 31.4×152

•°• Boiling point of solution (Tb) = Tb° + ∆Tb

= 56.30 + 0.16

= 56.46℃ ans

Answer 2

Explanation:

Boiling point of acetone = T= 56.30 °C = 329.30 K

Boiling point of the solution with 0.456 g of camphor = Tb

⚠️Tb=Tb-T

⚠️Tb=kf x m

Kf per 100g of acetone =17.2k g/mol

Kf per 1000g of acetone =1.72k kg/mol

⚠️Tb=1.72k kg/mol x 0.456g/152g/molx0. 0314kg =>0.164k

Tb=⚠️Tb+T

Tb=0.164k+329.30k=>329.464k

Tb=56.46°C

The boiling point of a solution is 56.46°C

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