Question

Maximum number of molecules is present in:

• A. 6 grams of helium
• B. 0.75 moles of nitrogen
• C. 44.8 lit of methane at STP
• D. 90 grams of glucose

Answer 1

Answer: (C)

44.8 lit of methane at STP

No. of molecules $=$ No. of moles $\times N_A$

Here, $N_A$ is the Avogadro number.

$6$ g of helium corresponds to $\frac{6}{4}=1.5$ moles or $1.5N_A$ molecules.

$0.75$ moles of nitrogen contains $0.75N_A$ molecules.

$44.8$ L of methane at STP corresponds to $\frac{44.8}{22.4}=2$ moles or $2N_A$ molecules.

$90$ g of glucose corresponds to $\frac{90}{180}=0.5$ moles or $0.5N_A$ molecules.

Hence, option C is correct.