Some Basic Concepts Of Chemistry Question 183
Question: Ratio of $ C _{p} $ and $ C _{v} $ of a gas X is 1.4, the number of atom of the gas ?X? present in 11.2 litres of it at NTP will be [CBSE 1999]
Options:
A) $ 6.02\times 10^{23} $
B) $ 1.2\times 10^{23} $
C) $ 3.01\times 10^{23} $
D) $ 2.01\times 10^{23} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{C _{P}}{C _{V}}=1.4 $ so, given gas is diatomic 11.2L $ =3.01\times 10^{23} $ molecules
$ \therefore $ No. of atoms $ =3.01\times 10^{23}\times 2 $ $ =6.023\times 10^{23} $ atoms
