Consider the following data.

(i) $2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}_2\text{(g)} + 1200~\text{kJ/mol}$

(ii) $\text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{HCl(g)} + 164~\text{kJ/mol}$

(iii) $\text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)} + 83~\text{kJ/mol}$

(iv) $\text{Al}_2\text{Cl}_6\text{(s)} + \text{aq} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 663~\text{kJ/mol}$

The enthalpy of formation of anhydrous solid $\text{Al}_2\text{Cl}_6$ is :

19.5 g of fluoro acetic acid (molar mass = 78 g mol⁻¹) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1^∘C. What is $K_a$ of fluoro acetic acid?

(For water, $K_f = 1.86$ K kg mol⁻¹). Assume molarity and molality to have same values.

The solubility product constants of $\mathrm{Ag_2CrO_4}$ and $\mathrm{AgBr}$ are $32x$ and $4y$ respectively at 298 K.

The value of $$\left( \frac{\text{molarity of } \mathrm{Ag_2CrO_4}}{\text{molarity of } \mathrm{AgBr}} \right)$$ can be expressed as :

An electrochemical cell is constructed using half cells in the direction of spontaneous change:

Fe(OH)₂(s) + 2e⁻ → Fe(s) + 2OH⁻(aq)      E^θ = −0.88 V

and AgBr(s) + e⁻ → Ag(s) + Br⁻(aq)      E^θ = +0.07 V

Which of the following option is correct?