Question

The normality of a $500$ ml $Ca(OH{)}_2$ solution which contains $9.033\times {10}^{22}$ hydroxide ions, is?

• A. $\frac{4}{1}N$
• B. $\frac{3}{10}N$
• C. $\frac{1}{5}N$
• D. $\frac{3}{7}N$

Answer 1

The correct option is B $\frac{3}{10}N$

Moloes of $O{H}^{-}=\frac{9.033\times {10}^{22}}{6.023\times {10}^{23}}=0.15$

$\therefore$ Moles of $Ca(OH{)}_2=\frac{1}{2}\times 0.15=0.075$

Molarity $=\frac{0.075}{500}\times 1000=0.15$

Molarity $=$ molarity $\times$ [$n$-$factor$ of $Ca(OH{)}_2$]

$=0.15\times 2=0.3=\frac{3}{10}N$