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

Options

A. `11.2ml`
B. `22.4ml`
C. `33.6ml`
D. `44.8ml`

Answer

4

Explanation

`{:(PbS,+,4H_(2)O_(2),rarr,PbSO_(4),+,4H_(2)O),(0.1" mole",,34xx4,,303" gms",,),(2.39" gms",,136" gms",,,,):}`
`2.39` gms `PbS ` require `136` gms of `H_(2)O_(2)`
We know `10V H_(2)O_(2)=3.035gms` of `H_(2)O_(2)`
Hence` 3.035"gms "H_(2)O_(2)=1ml` of `10` vol
`136 gms H_(2)O_(2)=(1)/(3.035)xx136ml=44.8ml`.