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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» <html><body><p></p>Solution :After hydrolysis, acid products are obtained. This suggests that the substanece is an acid <a href="https://interviewquestions.tuteehub.com/tag/chloride-915854" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORIDE">CHLORIDE</a>. 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 <a href="https://interviewquestions.tuteehub.com/tag/verified-2323104" style="font-weight:bold;" target="_blank" title="Click to know more about VERIFIED">VERIFIED</a> by the <a href="https://interviewquestions.tuteehub.com/tag/following-463335" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWING">FOLLOWING</a> way: <br/> `SO_(2)Cl_(2)(135g)=BaSO_(4)(233.4g)` <br/> `therefore0.4g-=[(233.4)/(135)xx0.4]g=0.692g` <br/> This agrees very nearly with the given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> `=0.7g` further, 135 g of `SO_(2)Cl_(2)-=H_(2)SO_(4)+2HCl` (by hydrolysis 4 <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a>). <br/> `1.53g-=(4)/(135)xx1.53` equivalent `=0.0453` equivalent <br/> `91 " mL of " 0.5 N NaOH-=(91)/(1000)xx0.5=0.0455` equivalent <br/> This also agrees with the given data. <br/> Therefore, the formula is `SO_(2)Cl_(2)`.</body></html>


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