A) 5.0 g
B) 1.25 g
C) 2.5 g
D) 4 g
Correct Answer: C
Solution :
\[\overset{+2\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{CuS{{O}_{4}}}}\,+2KI\to {{K}_{2}}S{{O}_{4}}+\overset{+2\,\,\,\,\,}{\mathop{Cu{{I}_{2}}}}\,\]; \[2\overset{+2\,\,\,\,\,}{\mathop{Cu{{I}_{2}}}}\,\to \overset{+1\,\,\,\,\,}{\mathop{Cu{{I}_{2}}}}\,+{{I}_{2}}\] \[{{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}\] Eq. wt. Of \[CuS{{O}_{4}}.5{{H}_{2}}O=\text{Mol}\text{. wt}\text{.}=\text{250}\] 100 ml of 0.1 N hypo º 100 ml of 0.1 N \[CuS{{O}_{4}}.5{{H}_{2}}O\] \[=\frac{250\times 0.1\times 100}{100}=2.5\,gm\]
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