Molarity  and  molality of 500 ml of a solution are 1M and 1.5m respectively. Weight of solvent present in the solution 

$1=\frac{weight.ofsolute}{mol.weight.soulute}\times \frac{1000}{500}\_\_\_\_\_\_\_\left(1\right)$

$1.5=\frac{weight.ofsolute}{mol.weight.soulute}\times \frac{1000}{wt.ofsolvent}\_\_\_\_\_\_\_\left(2\right)$

$\frac{1}{1.5}=\frac{2\times wt.ofsolvent}{1000}$

$\therefore$wf. Of solvent $=\frac{500}{1.5}=333.33gm$