The molal lowering of vapor pressure for `H_(2)O` at `100^(@)C` is

Explanation

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