Question

The number of gram molecule of oxygen in $6.02\times {10}^{24}$ CO molecules is :

• A. 10 gm molecules
• B. 5 gm molecules
• C. 1 gm molecules
• D. 0.5 gm molecules

Answer 1

Answer: (B)

5 gm molecules

Given :

$6.02\times {10}^{24}$ $CO$ molecules

We know,

$1$ mole $CO$ equivalent to $6.023\times {10}^{23}$ molecules of $CO$

$6.023\times {10}^{23}$ molecules of $CO$ contain $6.023\times {10}^{23}$ atoms of $O$

Then, $6.02\times {10}^{24}$ $CO$ molecules contain $6.02\times {10}^{24}$ atoms of $O$

$\Rightarrow$ no. of gram atoms of $O=\frac{6.02\times {10}^{24}}{6.02\times {10}^{23}}$

$=10gm$ atoms $⟶1$

As oxygen is a diatomic molecule.

No. of gm of molecules of oxygen=$\frac{10gmatoms}{2atoms/molecule}=5gm$ molecules

Therefore, the correct option is B.