Which of the following contains 6.02 times 10 | vedclass

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

Question

(A) The number of molecules is calculated using the formula: $	ext{Number of molecules} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Gi

Vedclass · Accepted Answer

$2.8 \ g \ 	ext{of} \ N_2$

Vedclass · Answer

(A) The number of molecules is calculated using the formula: $	ext{Number of molecules} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Given $N_A = 6.02 	imes 10^{23} \ 	ext{molecules/mol}$. We need $6.02 	imes 10^{22}$ molecules,which is $0.1 \ 	ext{mol}$. For $A$: $2.8 \ g \ N_2$ $(M = 28 \ g/mol)$ = $0.1 \ 	ext{mol}$. Since $N_2$ is a molecule,it contains $0.1 	imes 6.02 	imes 10^{23} = 6.02 	imes 10^{22}$ molecules. For $B$: $1.6 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.05 \ 	ext{mol}$. For $C$: $Na$ is an atom,not a molecule. For $D$: $1.4 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.04375 \ 	ext{mol}$. Thus,$A$ is the correct answer.

Which of the following contains $6.02 \times 10^{22}$ molecules?

Similar Questions

If one atom of an element $X$ weighs $6.643 \times 10^{-23} \ g$,then find the number of moles of atoms in $50 \ kg$ of element $X$. (in $moles$)

What is the number of moles and total number of atoms respectively present in $5.6 \ cm^{3}$ of ammonia gas at $STP$?

The largest number of molecules are in:

$A$ person adds $1.71 \ g$ of sugar $(C_{12}H_{22}O_{11})$ to sweeten their tea. How many carbon atoms are added? (Molar mass of sugar $= 342 \ g/mol$)

How many moles of helium gas occupy $22.4 \ L$ at $0^o C$ and $1 \ atm$ pressure?

Vedclass Test Series

Exam Paper Generator

Online Exam Module