A) 18
B) 68
C) 34
D) 17
Correct Answer: D
Solution :
Key Idea Equivalent weight \[=\frac{molecular\text{ }weight}{\begin{align} & change\text{ }in\text{ }oxidation\text{ }number\text{ }of\text{ }atoms \\ & present\text{ }in\text{ }one\text{ }molecule \\ \end{align}}\] In the following equation. \[{{H}_{2}}\overset{-2}{\mathop{S}}\,+NH{{O}_{3}}\xrightarrow{{}}2{{H}_{2}}O+2N{{O}_{2}}+\overset{o}{\mathop{S}}\,\] Increase in oxidation state = 2 hence, the equivalent weight of \[{{H}_{2}}S=\frac{molecular\text{ }weight}{change\text{ }in\text{ }oxidation\text{ }number}\] \[=\frac{37}{2}=17\]
You need to login to perform this action.
You will be redirected in
3 sec
