The volumes of 4N HCl and 10N HCl required to make 1L of 6 N HCl are

solutions mixed are 4N HCl and 10N HCl to form 1l of 6N HCl

Total volume of 6N solution = 1lit

4N HCl sol.            10N HCl sol.

V₁= v l                      V₂= (1-v) l

N₁= 4                        N₂=10

$\mathrm{Resultantnormality} ,\mathrm{N}=\frac{{\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}$

$6=\frac{\mathrm{v}\left(4\right)+(1-\mathrm{v})10}{1}$

 6= 4v+10-10v

 6v=4

$\mathrm{V}=\frac{2}{3}\mathrm{L}=0.67\mathrm{L}$

 V₁=0.67    V₂= 1-0.67 = 0.33 L