Mix Examples of Some Basic Concept of Chemistry Questions in English

(D) To find the number of atoms,we calculate the moles of molecules and multiply by the atomicity and Avogadro's number $(N_A)$:
$A$. $2 \text{ g}$ of $O_2 = 2/32 = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
$B$. $4 \text{ g}$ of $SO_2 = 4/64 = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 3 \times N_A = 0.1875 N_A$.
$C$. $1400 \text{ mL}$ of $O_2 = 1.4 \text{ L} / 22.4 \text{ L/mol} = 0.0625 \text{ mol}$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
$D$. $0.05 \text{ L}$ of $He = 0.05 / 22.4 \approx 0.00223 \text{ mol}$. Atoms = $0.00223 N_A$.
$E$. $0.0625 \text{ mol}$ of $H_2$. Atoms = $0.0625 \times 2 \times N_A = 0.125 N_A$.
Comparing the results,options $A, C$,and $E$ have the same number of atoms $(0.125 N_A)$.

(D) Step $1$: Calculate the number of atoms in $2$ moles of cyclohexane $(C_6H_{12})$.
One molecule of cyclohexane contains $6 + 12 = 18$ atoms.
Total atoms $= 2 \text{ mol} \times 18 \times N_A = 36 N_A$.
Step $2$: Calculate the number of atoms in $684 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$.
Molar mass of sucrose $= (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \text{ g/mol}$.
Moles of sucrose $= 684 \text{ g} / 342 \text{ g/mol} = 2 \text{ mol}$.
One molecule of sucrose contains $12 + 22 + 11 = 45$ atoms.
Total atoms $= 2 \text{ mol} \times 45 \times N_A = 90 N_A$.
Step $3$: Calculate the number of atoms in $90.8 \text{ L}$ of dihydrogen $(H_2)$ at $STP$.
Moles of $H_2 = 90.8 \text{ L} / 22.7 \text{ L/mol} \approx 4 \text{ mol}$ (using standard molar volume $22.7 \text{ L/mol}$ at $1 \text{ bar}$ $STP$).
One molecule of $H_2$ contains $2$ atoms.
Total atoms $= 4 \text{ mol} \times 2 \times N_A = 8 N_A$.
Comparing the values: $B (90 N_A) > A (36 N_A) > C (8 N_A)$.
Thus,the correct order is $B > A > C$.