Question

What will be the normality of a solution obtained by mixing $0.45N$ and $0.60NNaOH$ in the ratio $2:21$ by volume?

Answer 1

Let volume of $0.45N$ $NaOH=V$

$\therefore$ Volume of $0.60N$ $NaOH=\frac{21}{2}V$

Now,

$N_1{V}_1+N_2{V}_2=N_3{V}_3$

where, $N_1$, $N_2$ are the normality of $0.45N$ and $0.60N$ $NaOH$ respectively and $V_1$ and $V_2$ are their volumes respectively.

$N_3$ and $V_3$ are the normality & volume of the final solution.

$\therefore$ $(0.45\times V)+(0.60\times \frac{21}{2}V)=N_3\times (V+\frac{21}{2}V)$

$\therefore$ $0.45V+6.3V=N_3\times 11.5\times V$

$\therefore$ $0.45+6.3=N_3\times 11.5$

$\therefore$ $N_3=\frac{0.45+6.3}{11.5}$

$\therefore$ $N_3=0.587N$

$\therefore$ The normality of solution obtained is $0.587N$