Chemistry - Mole Concept Question with Solution | TestHub

$\underset{\frac{1.8}{\mathrm{M}}}{\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_{\mathrm{x}}{\mathrm{Cl}}_{\mathrm{y}}}+\underset{\frac{1.8}{\mathrm{M}}}{\mathrm{xHCL}}\to \mathrm{x}\underset{\mathrm{aq}}{{\mathrm{NH}}_4^{+}}+{\mathrm{Co}}^{\mathrm{y}+}+\underset{\frac{1.8(\mathrm{x}+\mathrm{r})}{\mathrm{M}}}{(\mathrm{x}+\mathrm{y}){\mathrm{Cl}}^{-}}$

$\frac{1\text{.}8\text{x}}{\text{M}}=20\times 1\text{.}54\times 10^{-3}=30\text{.}8\times 10^{-3}$........(I)

$\frac{1·8\left(\text{x}+\text{y}\right)}{\text{M}}=\frac{7\text{.}735}{143\text{.}5}=53\text{.}9\times 10^{-3}$........(II)

$\frac{\text{II}}{\text{I} }\Rightarrow \frac{53\text{.}9\times 10^{-3}}{30\text{.}8\times 10^{-3}}=1\text{.}75=\frac{\text{x}+\text{y}}{\text{x}}=1+\frac{\text{y}}{\text{x}}$

$\frac{\text{y}}{\text{x}}=0\text{.}75$

Substituting in (I)$\Rightarrow$

$\frac{1\text{.}8\text{x}}{59+17\text{x}+35\text{.}5\times \text{0}\text{.}75\text{x}}=30\text{.}8\times 10^{-3}$

$\therefore$x = 3.98 = 4$\therefore$y = 3