Question

$4$g $NaOH$ is added in $100$mL of $0.5$M $NaOH$ solution and solution was made $1$L with addition of water $20$mL of above solution can:

• A. Completely neutralise $20$mL of $0.4$M $H_2{C}_2{O}_4$ (oxalic acid)
• B. Completely neutralise $20$mL of $0.5$M $HN{O}_3$ solution
• C. Completely neutralise $15$mL of $0.2$M $HCl$ solution
• D. None of these

Answer 1

The correct option is C Completely neutralise $15$mL of $0.2$M $HCl$ solution

Molarity= $\frac{\text{Number of moles of solute}}{Volume}$

$0.5M$ $NaOH$ is $0.5$ moles of $NaOH$ dissolved in $1L$ of solvent

$100ml$ of $0.5M$ $NaOH=100\times 0.5=50mmol$ $NaOH$

Mass of $NaOH=50\times {10}^{-3}\times 40=2g$

$4g$ $NaOH$ is added to $100ml$ of $0.5M$ $NaOH$

$\therefore 6g$ of $NaOH$ is present

Molarity= $\frac{6/40}{100/1000}=1.5M$

$M_1{V}_1=M_2{V}_2$

$\therefore M_2=\frac{1.5\times 100}{1000}=0.15M$ (After dilution)

$\therefore (MV{)}_{NaOH}=0.15\times 20=3=15\times 0.2=(MV{)}_{HCl}$

$\therefore$ It completely neutralises $15ml$ of $0.2M$ $HCl$ solution.