The volume of water that must be added to a mixture of 250 ml of 0.6 M HCI and 750 ml of 0.2 M HCl to obtain 0.25 M solution of HCI is :

$\text{ No. of moles }=\text{ Molarity }\times \text{ volume of solution }$

$\text{ Initial Number of moles }=250\times 0.6+0.2\times 750$

$\text{ Final number of moles }=0.25\left(250+750+V_{\text{required }}\right)$

${\mathrm{V}}_{\text{required }}=\frac{0.6\times 250+0.2\times 750}{0.25}\left[1000\right]=200\mathrm{ml}$.