Question

Magnesium hydroxide is the white milky substance in milk of magnesia. What mass of $$Mg(OH)_{2}$$ is formed when $$15 ml$$ of $$ 0.2 M - NaOH $$ is combined with $$12$$ ml of $$ 0.15 M - MgCl_{2}?$$

Answer 1

Correct option is A. $$0.087 g$$
$$ NaOH :- 15 ml\, 0.2 M$$
$$ MgCl_{2} :- 12 ml \, 0.15 M $$
milliequivalent of $$ NaOH = 0.2 \times 15 = 3m.e.$$
millequivalent of $$ MgCl_{2} = 0.15 \times 2 \times 12 = 3.6 m.e.$$
So only 3 m.e. of NaOH and $$ MgCl_{2}$$ reacted to give
3 m.e. of $$Mg(OH)_{2}$$
Now,
Eq.ut of $$ Mg(OH)_{2} = \dfrac{58}{2} == 29 gm$$
$$ \therefore \dfrac{ut. of Mg(OH)_{2}}{29} \times 1000 = 3 $$
$$ \therefore $$ ut. of $$ Mg(OH)_{2} = 0.087 g $$