$(a)\;6\qquad(b)\;7\qquad(c)\;4\qquad(d)\;2$

Step 1:

$H_2O_2$ is marked 22.4 volume,which means 1litre $H_2O_2$ gives 22.4litre of $O_2$ at NTP.

$2H_2O_2\rightarrow 2H_2O+O_2$

$22400ml$ of $O_2$ is obtained from =68g$H_2O_2$

$22.4ml$ of $O_2$ is obtained from =$\large\frac{68\times 22.4}{22400}$

$\Rightarrow 0.068gH_2O_2$

Step 2:

This amount is present in 1ml $H_2O_2$ solution.

$H_2O_2+H_2S\rightarrow 2H_2O+S$

$34g H_2S$ is oxidized by =68g

$3.5g H_2S$ is oxidized by =$\large\frac{68\times 3.5}{34}$

$\Rightarrow 7g H_2O_2$

Hence (b) is the correct answer.

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