Chemistry - Concentration Terms Question with Solution | TestHub

${\text{M}}_{\text{eq}\text{.}}\text{of}{\text{CuSO}}_{\text{4}}\text{×}{\text{xH}}_{\text{2}}\text{O}\text{=}{\text{M}}_{\text{eq}\text{.}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}{\text{M}}_{\text{eq}\text{.}}\text{NaOH}$
${\text{M}}_{\text{eq}\text{.}}{\text{CuSO}}_{\text{4}}\text{=}{\text{M}}_{\text{eq}\text{.}}\text{NaOH}$
$\frac{{\text{10}}^{\text{3}}\text{×}\text{wt}\text{.}}{\text{Mol}\text{.}\text{mass}}\times \text{v}\text{.f}\text{.}=\text{N}\times \text{V}$
$\frac{1.245}{\text{M}}\times 2\times {10}^3=\frac{4}{20}\times 2$
$\text{M}=1.245\times 2\times {10}^3\times 10$
$\text{M}=249$
Molecular weight of$\mathrm{C}\mathrm{u}\mathrm{S}{\mathrm{O}}_4.\mathrm{x}{\mathrm{H}}_2\mathrm{O}=63.5+32+64+18\mathrm{x}$
$=159.5+18\mathrm{x}$
$\text{249 = 159}\text{.5 + 18x}$
So,$\text{x = 5}$.