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

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

Total charge of 'Cl' on LHS = -14
Total charge of 'Cl' on RHS = -8
Increases in Oxidation number/14mole of HCl = 6
MW of HCl = 36.5

${\mathrm{E}}_{\mathrm{HCl}}=\frac{14(36.5)}{6}=85.1$