The maximum number of molecules is present in

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the maximum number of molecules,we calculate the number of moles $(n)$ for each option:$A$: $n = \frac{0.5 \ g}{2 \ g/mol} = 0.25 \ mol$.

Vedclass · Accepted Answer

$15 \ L$ of $H_2$ gas at $STP$

Vedclass · Answer

(C) To find the maximum number of molecules,we calculate the number of moles $(n)$ for each option:$A$: $n = \frac{0.5 \ g}{2 \ g/mol} = 0.25 \ mol$. Molecules $= 0.25 	imes N_A = 0.25 N_A$.$B$: $n = \frac{10 \ g}{32 \ g/mol} = 0.3125 \ mol$. Molecules $= 0.3125 	imes N_A = 0.3125 N_A$.$C$: $n = \frac{15 \ L}{22.4 \ L/mol} \approx 0.67 \ mol$. Molecules $= 0.67 	imes N_A = 0.67 N_A$.$D$: $n = \frac{5 \ L}{22.4 \ L/mol} \approx 0.22 \ mol$. Molecules $= 0.22 	imes N_A = 0.22 N_A$.Comparing the values,$15 \ L$ of $H_2$ gas at $STP$ contains the maximum number of molecules.

Similar Questions

Which of the following statements is incorrect?

How many water molecules are there in one drop of water $($Volume $= 0.0018 \ mL)$ at room temperature?

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is $1 : 4$. The ratio of the number of their molecules is:

The number of water molecules in $1 \, L$ of water is = ............

The number of atoms present in a drop of water weighing $1.8 \ g$ is.....

Vedclass Test Series

Exam Paper Generator

Online Exam Module