`SO_(2)Cl_(2)` is dissolved in water to obtain `H_(2)SO_(4)` and HCL. In two different cases, excess `Ba(OH)_(2)` and excess `CH_(3)COOAg` were added to obtain precipitates having masses `m_(1)` and `m_(2)` grams respectively , starting with one mole of `SO_(2)Cl_(2)` in each case. Then (`Ba=137, Ag=108, S=32, Cl=35.5)`:-`

Options

A. `m_(1)=233g`
B. `m_(2)-m_(1)=54g`
C. `m_(2)=287g`
D. All of these

Answer

D

Explanation

`SO-(2)Cl_(2)+2H_(2)Orarr`

`m_(1)=1molxx233g//mol=233g`
`m_(2)=2molxx143.5g//mol=287g`