A Water technician recommends that you treat your well water that contain $1.7 mg H_{2} S$ per litre with $H_{2} O_{2}$ , in order to convert $H_{2} S$ to Sulphate (Which does not have an offensive smell or taste) What volume $(in ml)$ of a $3 %$ by mass solution of $H_{2} O_{2}$ is required to treat 1000 litres of your well water (Assume that no other species are present to react with $H_{2} O_{2}$ and the density of $H_{2} O_{2}$ solution is $1.0 g / ml .$ (atomic weight of S = 32, O = 16, H = 1)

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answer is 226.6.

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Detailed Solution

$H_{2} S + 4 H_{2} O_{2} \to H_{2} S O_{4} + 4 H_{2} O 34 136$

1000 litres of Well water contain $\frac{1.7 \times 1000}{1000} = 1.7 g$

To convert 34g of $H_{2} S$ $136 g oƒ H_{2} O_{2}$ required
To convert 1.7g of $H_{2} S$ the $H_{2} O_{2}$ required is $\frac{1.7 \times 136}{34} = 6.8 g$
Since $H_{2} O_{2}$ solution is $3 %$
3g of $H_{2} O_{2}$ is present in 100 ml
6.8g of $H_{2} O_{2}$ is present in $\frac{6.8 \times 100}{3}$ $= 226.7 m l$