Explanation

(a)
Suppose the volume of 6 M HCL required to obtain 1 L of 3 M
HCl = $x$ L
$\therefore$ volume of 2 N HCl required = ($1 - x$) L
Applying the molarity equation
$M_{1}V_{1} + M_{2}V_{2} = M_{3}V_{3}$
$6M\ HCl + 2\ MHCl\ \ \ \ 3M\ HCl$
$6x + 2(1 - x) = 3 \times 1$
$4x = 1$
$x = 0.25$ L
Hence, volume of 6M HCl required = 0.25 L
and volume of 2M HCl required $= 0.75$ L