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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    The weight of hypo\[(N{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O)\]required to make\[100\text{ }c{{m}^{3}}\]of 0.2 N solution for use in the reaction \[2{{S}_{2}}O_{3}^{2-}+{{I}_{2}}\xrightarrow{{}}{{S}_{4}}O_{6}^{2-}+2{{I}^{-}}\]will be

    A)  1.49 g                                  

    B)  2.98 g

    C)  4.96 g                                  

    D)  8.94 g

    Correct Answer: C

    Solution :

                     According to the given reaction \[2{{S}_{2}}O_{3}^{2-}\xrightarrow{{}}{{S}_{4}}O_{6}^{2-}+2{{e}^{-}}\] \[\therefore \]Eq. wt. of\[N{{a}_{2}}{{S}_{2}}{{O}_{2}}.5{{H}_{2}}O\]                 \[=\frac{mol.wt.}{1}=\frac{248}{1}=248\] \[1000\text{ }c{{m}^{3}}\]of 1 N solution will require \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}-5{{H}_{2}}O=248\text{ }g\] \[\therefore \]\[100\text{ }c{{m}^{3}}\]of 0.2 N solution will require \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O=\frac{248\times 0.2}{1000}\times 100=4.96\,g\]


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