Why is one molar aqueous solution more concentrated than one molal solution?

Show that the relative lowering of vapour pressure is given by:

$\frac{{\mathrm{p}}^0-\mathrm{p}}{{\mathrm{p}}^0}=\frac{{\mathrm{W}}_2/{\mathrm{M}}_2}{{ \mathrm{W}}_1/{\mathrm{M}}_1+{\mathrm{W}}_2/{\mathrm{M}}_2}$

where ${\mathrm{W}}_1$ and ${\mathrm{W}}_2$ are the weights and ${\mathrm{M}}_1$ and ${\mathrm{M}}_2$ are the molecular weights of solvent and solute respectively, ${\mathrm{p}}^0$ is the vapour pressure of pure solvent and $\mathrm{p}$ that of solution.