The volume of water that must be added to a mixture of $250\text{mL}$ of $0.6\text{M}HCl\text{and}750\text{mL}\text{of}0.2\text{M}HCl$ to obtain $0.25$ solution of HCl?

The correct option is D $1.2\text{L}$
$\text{Number of moles of}0.6\text{M}HCl=0.6\times 0.25=0.15\text{moles}$

$\text{Number of moles of}0.2\text{M}HCl=0.2\times 0.55=0.15\text{moles}$

$\Rightarrow \text{Total number of moles}=0.15+0.15=0.3\text{moles}$

$\text{Number of moles of}0.25\text{M}HCl=0.25\times V\left(\text{L}\right)=0.3\text{moles}$

$\Rightarrow V\left(\text{L}\right)=0.3\text{mol}/0.25{\text{mol L}}^{-1}$

$\Rightarrow V=1.2\text{L}$.