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$25 \mathrm{ml}$ of a solution of barrium hydroxide on titration with a $0.1$ molar solution of hydrochloric acid gave a litre value of $35 \mathrm{ml}$. The molarity of barium hydroxide solution was
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$0.07$
$0.07$
$\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCI} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}$
Applying Molarity equation, $\frac{\mathrm{M}_1 \mathrm{~V}_1}{\frac{1}{\left(\mathrm{Ba}(\mathrm{OH})_2\right)}}=\frac{\mathrm{M}_2 \mathrm{~V}_2}{2}$ or $25 \times \mathrm{M}_1=\frac{0.1 \times 35}{2} \therefore \mathrm{M}_1=\frac{0.1 \times 35}{2 \times 25}=\frac{0.7}{10}=0.07$
Applying Molarity equation, $\frac{\mathrm{M}_1 \mathrm{~V}_1}{\frac{1}{\left(\mathrm{Ba}(\mathrm{OH})_2\right)}}=\frac{\mathrm{M}_2 \mathrm{~V}_2}{2}$ or $25 \times \mathrm{M}_1=\frac{0.1 \times 35}{2} \therefore \mathrm{M}_1=\frac{0.1 \times 35}{2 \times 25}=\frac{0.7}{10}=0.07$
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