In transforming `0.01` mole of `PbS` to `PbSO_(4)`, the volume of '10 volume `H_(2)O_(2)` required will be `:`

Options

A. 11.2 mL
B. 22.4 mL
C. 33.6 mL
D. 44.8 mL

Answer

D

Explanation

`underset("(0.01 mole")(PbS(s))+underset((4xx0.01"mole"))(4H_2O_2(l))rarrPbSO_4+4H_2O(l)`
Wt. of `H_2O_2(l)` required for transforming 0.01 mole of `PbS = 4 xx 0.01 xx 34 ` g = 1.36 g
`2H_2O_2(l) rarr 2H_2O(l)+O_2(g)`
`2 xx 34 g " "` 22.4L at NTP
22.4L. `O_2` produced by `=2 xx34 g. H_2O_2`
10L. `O_2` produced by `=(2xx34)/(22.4)xx10g`
`=30.35g.H_2O_2`
So 1.L. of `H_2O_2` solution has = 30.35g. `H_2O_2`
1.36g. `H_2O_2` will be present in `=(1.36)/(30.35) ` L . `H_2O_2`
`=(1.36 xx1000)/(30.35) ` mL. `H_2O_2 = 44.8mL. H_2O_2`