Lowering of vapour pressure of an aqueous solution of a non-volatile, non-electrolyte 1 M aqueous solution at 100^oC is

1 M aqueous solution contains 1 mole of solute in 1000 g (55.5 moles) of water.
                       ${\mathrm{x}}_{\mathrm{B}}=\frac{1}{1+55.5}=\frac{1}{56.5}$
                                      ${\mathrm{p}}^{\mathrm{o}}=760 \mathrm{torr} \mathrm{at} {100}^{\mathrm{o}}\mathrm{C}$         
                    $\frac{∆\mathrm{p}}{{\mathrm{p}}^{\mathrm{o}}}={\mathrm{x}}_{\mathrm{B}},∆\mathrm{p}={\mathrm{p}}^{\mathrm{o}}{\mathrm{x}}_{\mathrm{B}}=\frac{760\times 1}{56.5}=13.45 \mathrm{torr}$