1.53 g of a compound containing only sulphur, oxygen and chlorine after easy hydrolysis with water yielded acid products which consumed 91 " mL of " `(N)/(2)` sodium hydroxide for complete neutralisation in a parallel experiment, 0.4 g of the compound after hydrolysis with water, was treated with excess of `BaCl_(2)` solution and 0.7 g of `BaSO_(4)` was precipitated. What is the formula of the compound?

Explanation

After hydrolysis, acid products are obtained. This suggests that the substanece is an acid chloride. Reaction with `BaCl_(2)` to yield `BaSO_(4)` implies that `H_(2)SO_(4)` is one of the product. The substance is (by surmise) `SO_(2)Cl_(2)` This can be verified by the following way:
`SO_(2)Cl_(2)(135g)=BaSO_(4)(233.4g)`
`therefore0.4g-=[(233.4)/(135)xx0.4]g=0.692g`
This agrees very nearly with the given data `=0.7g` further, 135 g of `SO_(2)Cl_(2)-=H_(2)SO_(4)+2HCl` (by hydrolysis 4 equivalents).
`1.53g-=(4)/(135)xx1.53` equivalent `=0.0453` equivalent
`91 " mL of " 0.5 N NaOH-=(91)/(1000)xx0.5=0.0455` equivalent
This also agrees with the given data.
Therefore, the formula is `SO_(2)Cl_(2)`.