Question

The increasing order of molarity with $25\, g$ each of $NaOH$, $LiOH , Al ( OH )_{3}, KOH , B ( OH )_{3}$ in same volume of water?

• A. $Al ( OH )_{3}< B ( OH )_{3}< KOH < NaOH < LiOH$
• B. $LiOH < NaOH < KOH < B ( OH )_{3}< Al ( OH )_{3}$
• C. $LiOH < NaOH < B ( OH )_{3}< KOH < Al ( OH )_{3}$
• D. $NaOH < LiOH < B ( OH )_{3}< Al ( OH )_{3}< KOH$

Answer 1

Answer: (A)

Molarity $(M)=\frac{\text { No. of moles of solute }}{\text { Volume of solution in litres }}$
Molarity $\propto n_{\text {solute }}$
$n_{ NaOH }=\frac{25}{40}=0.625$
$n_{ LiOH }=\frac{25}{24}=1.04$
$n_{ Al ( OH )_{3}}=\frac{25}{(17+3 \times 17)}=0.32 $
$n_{ KOH }=\frac{25}{(39+17)}=0.45$
$n_{ B ( OH )_{3}}=\frac{25}{(11+17 \times 3)}=0.403$