100 ml of 1M HCl, 200 ml of 2M HCl and 300 ml of 3M HCl are mixed with enough water to get 1M solution. The volume of water to be added is (in ml)

 The quantities provided in the question are:

${\mathrm{M}}_1=1\mathrm{M}, {\mathrm{V}}_1=100 \mathrm{ml}$

${\mathrm{M}}_2=2\mathrm{M}, {\mathrm{V}}_2=200 \mathrm{ml}$

${\mathrm{M}}_3=3\mathrm{M}, {\mathrm{V}}_3=300 \mathrm{ml}$

The product of total mass and volume can be expressed as:

${\mathrm{M}}_{\mathrm{total}}{\mathrm{V}}_{\mathrm{total}}={\mathrm{M}}_1{\mathrm{V}}_1+{\mathrm{M}}_2{\mathrm{V}}_2+{\mathrm{M}}_3{\mathrm{V}}_3$

$1\times {\mathrm{V}}_{\mathrm{total}}=1\times 100+2\times 200+3\times 300$

${\mathrm{V}}_{\mathrm{total}}=1400 \mathrm{ml}$

Volume of water to be added $={\mathrm{V}}_{\mathrm{total}}-({\mathrm{V}}_1+{\mathrm{V}}_2+{\mathrm{V}}_3)$

$=1400-(100+200+300)=800 \mathrm{ml}$

Hence, the required answer is C) 800.