A) \[\frac{\text{molecuar weight}}{\text{3}}\]
B) \[\frac{\text{molecuar weight}}{6}\]
C) \[\frac{\text{molecuar weight}}{1}\]
D) \[\frac{\text{molecuar weight}}{2}\]
Correct Answer: B
Solution :
\[{{K}_{2}}\overset{+6}{\mathop{C}}\,{{r}_{2}}{{O}_{7}}+4{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}+\overset{+3}{\mathop{C}}\,{{r}_{2}}(S{{O}_{4}}){{ & }_{3}}\] \[+\,4{{H}_{2}}O\,\,+\,\,3O\] Equivalent weight \[\text{= }\frac{\text{Molecular}\,\text{weight}}{\text{Chage}\,\text{in}\,\text{oxidiaton number}}\] Of atoms present in one molecule \[\therefore \] Equivalent weight \[\text{=}\,\frac{\text{Molecular}\,\text{weight}}{\text{2}\,\,\text{ }\!\!\times\!\!\text{ }\,\text{3}}\] \[\text{= }\frac{\text{Molecular}\,\text{weight}}{\text{6}}\]
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