• The boiling point of water was lowerd by 0.19°C by dissolving 0.152 g of a substance in 25g of water.Calculate the molecular weight of the substance .Molar freezing point depression constant for water is 18.5/100g.(ans 59.2)

ΔTb = Kbm

Kb = 18.5 /100g = 1.85 (per 1000g) given

ΔTb = increase in boiling point (given) = 0.190C

Wsolute of solute (given) = 0.152 g

W solvent = 25 g

Now  ΔTb = Kb (Wsolute/M.Mass of solute)(1000/ W solvent)

Thus 0.19 = 1.85 X(0.152/M)(1000/25)

M = 59.2 g

 Note- you have given molal depression constant but as question is based on elevation in boiling point. so, it will be molal elevation constant. 

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what is ur question..?

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 u couldnt see my question?but i had typd it

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 hey ..u need malar boiling point elevation const. to calculate this ques.. u hav mentioned molar freezing depression const. mind ur ques..

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u hav also mentioned that boiling point has depressed by adding non volatile substance.. what a hell is it?

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then why the hell are you showing ur anger to me..the question is only thet that !

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