1.

The vapour pressure of benzene at 25^(@)C is 639.7mmHg and the vapour pressure of solution of a solute in benzene at the same temperature is 631.9mm of mercury. Calculate the molality of the solution

Answer»

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SOLUTION :We have,
`(p^(0)-p)/(p^(0))=(n_(1))/(n_(2))` ……………….(Eqn. 1)
or `(n_(1))/(n_(2))=("MOLES of solute")/("moles of solvent")=(639.7-631.9)/(639.7)=0.0122`
Thus 1 mole of solvent `(C_(6)H_(6))` contains `0.0122` mole of solute, or `78g` of solvent`(C_(6)H_(6))` contains `0.0122` mole of solute `(C_(6)H_(6)=78)`
`:.` MOLALITY `=(0.0122)/(78)xx1000=0.156m`


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