Question

$3.5g$ of a mixture of $NaOH$ and $KOH$ was dissolved to form a solution and is made up to $250ml$. $25mL$ of this solution was completely neutralised by $17mL$ of $\frac{N}{2}HCl$ solution. Then, the percentage of $KOH$ in the mixture is:

• A. $80$
• B. $10$
• C. $34$
• D. $56$

Answer 1

The correct option is B $10$

Let $xg$ of $NaOH$ & $yg$ of $KOH$ be present in the mixture.
Thus, moles of $(NaOH+KOH)$ in the mixture $=\frac{x}{40}+\frac{y}{56}$
And, Molarity of the solution made of $3.5gofNaOH\&KOH$
$=\frac{\frac{7x+5y}{280}}{250}\times 1000$
$=[\frac{(7x+5y)}{280}\times 4]M$
$25mL$ of this solution is completely neutralised by $17mL\frac{N}{2}HCl(\frac{N}{2}=\frac{M}{2}\text{for HCl})$
Thus from,
$M_1{V}_1=M_2{V}_2$
$M_1\times 25=\frac{1}{2}\times 17\Rightarrow M_1=\frac{17}{50}=0.34M$
$\text{So,}\frac{7x+5y}{280}\times 4=0.34\&x+y=3.5$
$\Rightarrow \frac{7x+5(3.5-x)}{280}\times 4=0.34$
$\frac{7x+17.5-5x}{280}\times 4=0.34$

$\Rightarrow x=3.15g$ And, $y=3.5-3.15=0.35g$

Thus, mass of $NaOH$ in the mixture$=3.15g$
And, mass of $KOH=0.35g$
$\therefore$ $\%$ of $KOH$ in the mixture $=\frac{0.35}{3.5}\times 100=10\%$

Hence the correct answer is option (b).