What is the mass ratio of ethylene glycol ($\mathrm{C_2H_6O_2}$, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?

For 500 g of 0.25 molal aqueous solution, <br/><br/>0.25 = ${{{{{w_1}} \over {62}}} \over {{{500} \over {1000}}}}$ = ${{2{w_2}} \over {62}}$ ......(1) <br/><br/>For 250 mL of 0.25 molar aqueous solution <br/><br/>${{{{{w_2}} \over {62}}} \over {{{250} \over {1000}}}} = {{4{w_2}} \over {62}}$ .......(2) <br/><br/>Dividing equation (1) by (2), we get <br/><br/>${{0.25} \over {0.25}} = {{{{2{w_1}} \over {62}}} \over {{{4{w_2}} \over {62}}}}$ <br/><br/>$\Rightarrow {{2{w_1}} \over {62}} = {{4{w_2}} \over {62}}$ <br/><br/>$\Rightarrow {{{w_1}} \over {{w_2}}} = {4 \over 2} = {2 \over 1}$ <br/><br/><b>Note :</b> <p>(1) Molality of solution is defined as number of moles of solute present per kg of solvent.</p> <p>$m = {{no.\,of\,moles\,(solute)} \over {weight(in\,kg) of solvent}}$</p> <p>(2) Molarity of solution is defined as number of moles of solute present per litre of solution.</p> <p>$M = {{no.\,of\,moles\,(solute)} \over {v(in\,litre\,of\,solution)}}$</p>