25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydro chloric acid gave a titre value of 35 ml. The molarity of barium hydroxide solution was

$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2+2\mathrm{HCl}\to {\mathrm{BaCl}}_2+{\mathrm{H}}_2\mathrm{O}$

$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2 \mathrm{HCl}$

${\mathrm{M}}_1=? {\mathrm{M}}_2=0.1\mathrm{M}$

${\mathrm{V}}_1=25 \mathrm{mL} {\mathrm{V}}_2=35\mathrm{mL}$

${\mathrm{n}}_1=1 {\mathrm{n}}_2=2$

$\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{M_2{V}_2}{n_2}$

${\mathrm{M}}_1=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}\times \frac{{\mathrm{n}}_1}{{\mathrm{V}}_2}$

$=\frac{0.1\times 35}{2}\times \frac{1}{25}=\frac{7}{100}=0.07$