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In the reaction: `Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`, the equivalent weight of `Na_(2)S_(2)O_(3)` will be: (M= molecular weight of `Na_(2)S_(2)O_(3))`A. (a)`M//4`B. (b)`M//8`C. (c )`M//1`D. (d)`M//2` |
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Answer» Correct Answer - B `Na_(2) overset(+2)(S_(2) O_(3)) rarr Na_(2) overset(+6)(S O_(4))` the total change in oxidation number `=4xx2=8` `:. E_(Na_(2)S_(2)O_(3))=(mol. Wt.)/(v.f)=M/8` |
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