Question

$3.5\, g$ of a mixture of $NaOH$ and $KOH$ were dissolved and made up to $250\, mL\, .25\, mL$ of this solution were completely neutralised by $17\, mL$ of $( N / 2) HCl$ solution. Then, the percentage of $KOH$ in mixture is

• A. 80
• B. 10
• C. 34
• D. 56

Answer 1

Answer: (B)

Suppose the amount of $NaOH$ in the mixture $=x g$
$\therefore $ Amount of $KOH =(3.5-x) g$
$25\, mL$ of alkali solution $\equiv 17 mL$ of $( N / 2) HCl$ or, $250\, mL$ of alkali solution
$\equiv 170\, mL$ of $( N / 2) HCl$
$1000\, mL$ of $( N ) HCl$ contain $36.5\, g$ of $HCl$
$170\, mL$ of $N / 2 HCl$ will contain
$\frac{36.5 \times 170}{1000 \times 2}=3 \cdot 1025\, g$ of $HCl$
Eq. mass of $NaOH =40$ and that of $KOH =56$
$40\, g$ of $NaOH$ require $36.5\, g$ of $HCl$
$x g$ of $NaOH$ will require $\frac{36.5 \times x}{40} g$ of $HCl$
Similarly, $56\, g$ of KOH require $36.5\, g$ of HCl $(3.5-x) g$ of $KOH$ will require $\frac{36.5(3.5-x)}{56} g$ of $HCl$
Thus, $\frac{36.5 x}{40}+\frac{36.5(3.5-x)}{56}$
$=3.1025\, g$ of $HCl$
Or $x=3.15\, g =$ Amount of $NaOH$ in the mixture
$\therefore $ Amount of $KOH =(3 \cdot 5-3 \cdot 15) g =0 \cdot 35 g$
$3.5\, g$ of the mixture contain $0.35\, g$ of $KOH$
$100\, g$ of the mixture contain $=\frac{0 \cdot 35\, g \times 100}{35}$
$=10\, g$ of $KOH$
Hence, $\%$ of $KOH =10$