Ncert Solutions Chemistry Class 12th

Phosphinic acid is hypo phosphorous acid $\left({\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2\right)$ .

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+{\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2?{\mathrm{N}\mathrm{a}\mathrm{H}}_2{\mathrm{P}\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

For neutrization

${\left({\mathrm{N}}_1{\mathrm{V}}_1\right)}_{\text{acid}}={\left({\mathrm{N}}_2{\mathrm{V}}_2\right)}_{\text{base}}$

$0.1*10=0.1*{\left({\mathrm{V}}_{\mathrm{m}\mathrm{L}}\right)}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}$

For a given reaction, rate $=-\frac{1}{2}\frac{d{n}_A}{dt}=-\frac{1}{3}\frac{d{n}_B}{dt}=-\frac{2}{3}\frac{d{n}_C}{dt}$

$\text{rate}=\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{A}}}{\mathrm{d}\mathrm{t}}=\frac{2}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}=\frac{4}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{C}}}{\mathrm{d}\mathrm{t}}$

Due to inter molecular $\mathrm{H}$ -Bonding in $\mathrm{B}$ , than $\mathrm{A},\mathrm{B}$  is more soluble and having more B.P point than A.

$P_{\text{gas}}=\frac{{\mathbf{n}}_{\text{gas}}RT}{V}$

as $n,T$ & $V$  constant So

${\mathrm{P}}_{{\mathrm{H}}_2}={\mathrm{P}}_{{\mathrm{O}}_2}={\mathrm{P}}_{{\mathrm{H}}_{\theta }}=2\mathrm{a}\mathrm{t}\mathrm{m}$

So, ${\mathrm{P}}_{\text{Total}}={\mathrm{P}}_{{\mathrm{H}}_2}+{\mathrm{P}}_{{\mathrm{O}}_2}+{\mathrm{P}}_{\mathrm{H}\mathrm{e}}=6\mathrm{a}\mathrm{t}\mathrm{m}$   

Conc. ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$  acts as dehydrating agent.

Molar mass of given complex $=266.5\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .

On treating with conc. ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$  the mass

lost by the complex $==\frac{13.5}{100}\left(266.5\right)\approx 36\mathrm{g}=2$ moles of ${\mathrm{H}}_2\mathrm{O}$

Formula of the complex $=\left. [\mathrm{C}\mathrm{r}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{C}\mathrm{l}}_2\right]\mathrm{C}\mathrm{l}.2{\mathrm{H}}_2\mathrm{O}$