Analytical Chemistry Question 238
The equivalent weight of $ Zn{{( OH )}_2} $ in the following reaction is equal to its molecular weight divided by the number of electrons gained or lost in the reaction.
[Zn{(OH)}_2 + NO_3~Zn(OH)NO_3 + H_2O] $ [MH CET 1999]
Options:
A) $ \frac{Formula\ wt\text{.}}{2} $
B) $ \frac{Formula\ weight}{1} $
C) $ 3\times formula\ wt\text{.} $
D) $ 2\times formula\ wt\text{.} $
Show Answer
Answer:
Correct Answer: B
Solution:
Equivalent weight of $ Zn{{(OH)}_2}=\frac{Molecularweight}{acidity}=\frac{M}{2} $ Acidity of $ Zn{{(OH)}_2}=2, $ two $ OH $ groups are replaced.
BETA
