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Molal lowering of vapour presure of liqu...

Molal lowering of vapour presure of liquid is 1.008 mm Hg at `25^(@)C` in very dilute solution. The vapour pressure of liquid at `25^(@)C` is `x xx 10` mm Hg. The value of `x` is (molecular mass of liquid is `"18 g mol"^(-1)`)

Text Solution

Verified by Experts

The correct Answer is:
6
`m=(p_(0)-p)/(p_(0))xx(1000)/(m_(A))`
`1=(1.08)/(p_(0))xx(1000)/(18)`
`p_(0)=60`
i.e., `x xx 10=60`
`x=6`]
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