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Asked Nov 20, 2019
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What is the boiling point of a solution containing 1.50 g C9H8O4 in 75 grams CHCl3? (kCHCl3 = 3.63 oC/m; T*b= 61.7oC)

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Given,

Solution containing 1.50 g C9H8O4 in 75 grams CHCl3

kb CHCl3 = 3.63 oC/m

Tb= 61.7oC

Molar mass of C9H8O4  ...

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Molality of the solution is given below, W1000 Molality= М, W, А 1.5 1000 Molality 180.158 75 g Molality=0.111

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