An aqueous solution of a non-volatile solute boils at $100.17^{\circ} \mathrm{C}$. The temperature at which this solution will freeze (in ${ }^{\circ} \mathrm{C}$ ) is
$$ \begin{aligned} & \left(K_b\left(\mathrm{H}_2 \mathrm{O}\right)=0.512^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1},\right. \\ & \left.K_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1}\right) \end{aligned} $$
A possible mechanism for the gaseous reaction $2 \mathrm{H}_2+2 \mathrm{NO} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2$ is
Step 1:2 $\mathrm{NO} \rightleftharpoons \mathrm{N}_2 \mathrm{O}_2$
Step 2 : $\mathrm{N}_2 \mathrm{O}_2+\mathrm{H}_2 \longrightarrow \mathrm{~N}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O}$ (slow)
Step 3: $\mathrm{N}_2 \mathrm{O}+\mathrm{H}_2 \longrightarrow \mathrm{~N}_2+\mathrm{H}_2 \mathrm{O}$
The rate law for this reaction is
The reduction potential of a half-cell consisting of a Pt electrode immersed in $2.0 \mathrm{M} \mathrm{Fe}^{2+}$ and $0.02 \mathrm{M} \mathrm{Fe}^{3+}$ solution (in V) is
Given : $\left(\frac{2.303 R T}{F}=0.059, E_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}\right)$
The sol prepared by Bredig's arc method is $X$ and the charge of sol particles of it is $q . X$ and $q$ are respectively
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