10ml of a mixture of $H_2 and O_2$ is exploded.  If the final volume of gas becomes 1ml, then the composition of original mixture may be,

$H_2+1/2O_2\to H_{2}O$

1v       $1/2^v$

$\text{ Left: } x 1/2 x$

$\underset{10ml}{\underbrace{H_2\&O_2}}$

$\text{ Let vol. of }{H}_2\text{ is x }$

$\text{ Then vol. of }{O}_2\text{ is }10-x$

$V_{\text{Final }}=V_{\text{initial }}-(x+1/2x)$

$\begin{array}{l}1=10-(3/2x)\\ \frac{3}{2}x=9\end{array}$

$\therefore x=6ml$

$\therefore H_2\text{ and }{O}_2\text{ volumes are }x\text{ and }(10-x)$

=    6    and    (10 – 6 )

=    6   and    4