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?

1.53 g of a compound containing only sulphur, oxygen and chlorine afte

1.	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?
Answer» Solution :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)`.