Borax in water gives

${\mathrm{B}}_4{\mathrm{O}}_7^{2-}+7{\mathrm{H}}_2\mathrm{O}\stackrel{ }{\to }4{\mathrm{H}}_3{\mathrm{BO}}_3+2{\mathrm{OH}}^{-}$

How many grams of Borax $\left({\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7.10{\mathrm{H}}_2\mathrm{O}\right)$ are required to prepare 50 ml of 0.2 M solution.

The n-factor or valency factor of borax in the above reaction is 2

Meq. of borax in 50 ml solution = VN = 50 x 0.2 x 2 = 20

Let weight of borax required = W

   $\frac{\mathrm{W}}{\mathrm{E}}=\frac{\mathrm{NV}}{1000}$

$\begin{array}{l}\frac{\mathrm{W}}{382}\times 2\times 1000=20\\ \therefore {\mathrm{W}}_{\left({\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7.10{\mathrm{H}}_2\mathrm{O}\right)}=3.82\end{array}$