Hydrogen And Its Compounds Question 132
Question: The volume of oxygen liberated from $ 0.68gm $ of $ H_2O_2 $ is
[Pb. PMT 2004]
Options:
A) $ 112ml $
B) $ 224ml $
C) $ 56ml $
D) $ 336ml $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ 2H_2O_2\xrightarrow{{}}2H_2O+O_2 $
$ 2\times 34g $
$ 22400ml $
$ \because $
$ 2\times 34gm=68gm $ of $ H_2O_2 $ liberates $ 22400mlO_2 $ at STP
$ \therefore $
$ .68gm $ of $ H_2O_2 $ liberates $ =\frac{.68\times 22400}{68}=224ml $
BETA
