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Q.
$20$ volumes $H_2O_2$ solution has a strength of about
Hydrogen
Solution:
$\because 22.4$ litres $O_2$ at N.T.P. obtained by $68\, g$ of $H_2O_2$
$\therefore 1 $ litre $O_2$ at N.T.P. obtained by $\frac{68}{22.4} g$ of $H_2O_2$
$\therefore 20$ litres $O_2$ at N.T.P. obtained by
$\frac{68}{22.4} \times 20\,g$ of $H_2O_2 = 60.71\,g$ of $H_2O_2$
$\therefore 1000 \,ml \,O_2$ at N.T.P. obtained by $= 60.71 \,g$ of $H_2O_2$
$\therefore 100 \,ml \, O_2$ at N.T.P. obtained by
$= \frac{60.71}{1000} \times 100 = 6.71\%$