1.

Calculate the molal lowering of vapour pressure for H_(2)O at 100^(@)C.

Answer»

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Solution :Molal lowering of vapour pressure is the lowering of vapour pressure of water when 1 mole of the SOLUTE is dissolved in `1000g` of the SOLVENT `(H_(2)O)`. Further, vapour pressure of pure water `(p^(0))` at `100^(@)C` will be `760MM` as `100^(@)C` is its boiling point.
Now, we have,
lowering of v.p. `=p^(0)-p=p^(0)((n)/(n+N))`
`=p^(0)((n)/(N))`
`=760xx(1)/(1000//18)((n=1),(N=(1000)/(18)))`
`=13.68`mm


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