$\mathrm{\Delta H}°$  for a reaction ${\mathrm{F}}_2+2\mathrm{HCl}\to 2\mathrm{HF}+{\mathrm{Cl}}_2$ is given as –352.8 KJ. ${\mathrm{\Delta H}}_{\mathrm{f}}°$ for HF is $268.3KJmo{l}^{-1}$, then ${\mathrm{\Delta H}}_{\mathrm{f}}°$ of HCl would be

Given data: $\Delta H_f°$ of HF = $268.3KJmo{l}^{-1}$

$\Delta H°$  for a reaction ${\mathrm{F}}_2+2\mathrm{HCl}\to 2\mathrm{HF}+{\mathrm{Cl}}_2$ is –352.8 KJ.

Formula/Concept used: Hess’s law

Detailed solution: The reaction of formation of HF will be:

$\frac{1}{2}{\mathrm{H}}_2+\frac{1}{2}{\mathrm{F}}_2\to \mathrm{HF};\mathrm{ }\mathrm{\Delta H}=-268.3\mathrm{KJ}$…………..(i) 

${\mathrm{F}}_2+2\mathrm{HCl}\to 2\mathrm{HF}+{\mathrm{Cl}}_2;\mathrm{ }\mathrm{\Delta H}=-352.8\mathrm{KJ}$……………(ii)

The formation of HCl can be written by:

$\left(i\right)-\left(ii\right)\times \frac{1}{2}$ 

The enthalpy f the reaction will be:

$$\begin{gathered} \mathrm{\Delta H}=-268.3-\left(\frac{-352.8}{2}\right) \\ \mathrm{\Delta H}=-268.3+176.4 \\ \mathrm{\Delta H}=-91.9 {\mathrm{KJmol}}^{-1} \end{gathered}$$

Therefore, the correct option is C.