Question

The increasing order of molarity with $25g$ 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 $\propto n_{\text{solute }}$
$n_{NaOH}=\frac{25}{40}=0.625$
$n_{LiOH}=\frac{25}{24}=1.04$
$n_{Al(OH{)}_3}=\frac{25}{(17+3\times 17)}=0.32$
$n_{KOH}=\frac{25}{(39+17)}=0.45$
$n_{B(OH{)}_3}=\frac{25}{(11+17\times 3)}=0.403$