The equivalent weight of HCl in the given reaction K₂Cr₂O₇ + 14HCl $\to$ 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O is

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+14\stackrel{+1}{\mathrm{H}}\stackrel{-1}{\mathrm{Cl}}\to 2\mathrm{k}\stackrel{-1}{\mathrm{Cl}}+2\mathrm{Cr}{\stackrel{-1}{\mathrm{Cl}}}_3+3\stackrel{0}{{\mathrm{Cl}}_2}+7{\mathrm{H}}_2\mathrm{O}$

n-factor represents the number of H⁺ ions an acid can give or the number of OH⁻ ions a base can produce.
n-factor = ${\displaystyle \frac{6}{14}}$
As we know,

The molecular mass of HCl =$36.5 \mathrm{gm} /\mathrm{mole}$
Equivalent weight =$\frac{\mathrm{molecular} \mathrm{mass}}{\mathrm{n}- \mathrm{factor}}$
By putting the values in the above formula, we get
Equivalent weight = ${\displaystyle \frac{36.5\times 14}{6}}$
Equivalent weight = $85.1$

Hence, the equivalent weight of HCl in the given reaction is $85.1$.