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H(2)O(2) is marked 22.4 volume. How much...

`H_(2)O_(2)` is marked `22.4` volume. How much of it is required to oxidise `3.5 g H_(2)S` gas?

A

`10 mL`

B

`70 mL`

C

`100 mL`

D

`1000 mL`

Text Solution

Verified by Experts

The correct Answer is:
C
Density of 'x volume' `H_(2) O_(2)= (68 xx)/(22400)g //mL`
Density of '22.4 volume ' =`(68 xx 22.4)/(22400)= 0.068 g//"mol"`
`underset(34 g)(H_(2)S) + underset(68 g)(H_(2)O_(2)) to 2 H_(2) O+S`
34 g `H_(2)O_(2)` is oxidised by = `68 g H_(2)O_(2)`
3.4 g `H_(2)O_(2)` is oxidised by = `(68 xx 22.4)/(34) = 6.8 g H_(2)O_(2)`
0.068 g `H_(2)O_(2)` is present in = 1mL
`therefore 6.8 g H_(2)O_(2) ` in present in = `100 mL`
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