The mole concept Questions in English

(A) Calculate the mass of each substance:
$I$. $1$ mole of $N_2 = 1 \times 28 \ g = 28 \ g$
$II$. $0.5$ mole of $O_3 = 0.5 \times 48 \ g = 24 \ g$
$III$. $3.011 \times 10^{23}$ molecules of $O_2 = 0.5$ mole of $O_2 = 0.5 \times 32 \ g = 16 \ g$
$IV$. $0.5$ gram atom of $O = 0.5 \times 16 \ g = 8 \ g$
Comparing the masses: $8 \ g < 16 \ g < 24 \ g < 28 \ g$.
Thus,the order of increasing mass is $IV < III < II < I$.

(A) Volume of one drop $= (\frac{1}{25}) \ mL$
$\therefore$ Mass of $1$ drop $= V \times d = (\frac{1}{25} \ mL)(1 \ g / mL) = \frac{1}{25} \ g$
Number of moles of $H_2O = \frac{\text{Mass of water in one drop}}{\text{Molar mass of water}} = \frac{1/25}{18} = \frac{1}{450} \ mol$
$\therefore$ Number of $H_2O$ molecules $= \frac{1}{450} \ N_0 = \frac{0.02}{9} \ N_0$

(D) To find the number of atoms,we use the formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times \text{atomicity} \times N_{A}$.
For $A$: $1 \ g$ of $Ag$ (atomic mass = $108 \ g/mol$): $\text{Atoms} = \frac{1}{108} \times 1 \times N_{A} \approx 0.0092 \ N_{A}$.
For $B$: $1 \ g$ of $Fe$ (atomic mass = $56 \ g/mol$): $\text{Atoms} = \frac{1}{56} \times 1 \times N_{A} \approx 0.0178 \ N_{A}$.
For $C$: $1 \ g$ of $Cl_{2}$ (molar mass = $71 \ g/mol$): $\text{Atoms} = \frac{1}{71} \times 2 \times N_{A} \approx 0.0281 \ N_{A}$.
For $D$: $1 \ g$ of $Mg$ (atomic mass = $24 \ g/mol$): $\text{Atoms} = \frac{1}{24} \times 1 \times N_{A} \approx 0.0416 \ N_{A}$.
Comparing the values,$1 \ g$ of $Mg$ has the largest number of atoms.

(D) Mass of one electron = $9.108 \times 10^{-31} \ kg$.
Number of electrons in $1 \ mole = 6.023 \times 10^{23}$.
Mass of $1 \ mole$ of electrons = $(9.108 \times 10^{-31}) \times (6.023 \times 10^{23}) \ kg = 9.108 \times 6.023 \times 10^{-8} \ kg$.
Number of moles that weigh $1 \ kg = \frac{1}{9.108 \times 6.023 \times 10^{-8}} = \frac{1}{9.108 \times 6.023} \times 10^8 \ moles$.

(C) Given weight ratio of $CH_4$ and $SO_2$ is $1:2$.
Let the weight of $CH_4 = w$ and weight of $SO_2 = 2w$.
Molar mass of $CH_4 (M_1) = 16 \ g/mol$.
Molar mass of $SO_2 (M_2) = 64 \ g/mol$.
Number of moles of $CH_4 (n_1) = \frac{w}{16}$.
Number of moles of $SO_2 (n_2) = \frac{2w}{64} = \frac{w}{32}$.
The ratio of the number of molecules is equal to the ratio of the number of moles.
Ratio $= \frac{n_{SO_2}}{n_{CH_4}} = \frac{w/32}{w/16} = \frac{16}{32} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(C) $4.25 \text{ g}$ $NH_3 = (4.25 / 17) \times 4 \times N_A = 1 \times N_A$ atoms.
$(B)$ $8 \text{ g}$ $O_2 = (8 / 32) \times 2 \times N_A = 0.5 \times N_A$ atoms.
$(C)$ $2 \text{ g}$ $H_2 = (2 / 2) \times 2 \times N_A = 2 \times N_A$ atoms.
$(D)$ $4 \text{ g}$ $He = (4 / 4) \times 1 \times N_A = 1 \times N_A$ atoms.
Comparing the values,$2 \text{ g}$ $H_2$ contains the maximum number of atoms.

(C) To find the number of atoms,we calculate the moles of each substance and multiply by the number of atoms per molecule/formula unit and Avogadro's number $(N_{A})$.
$A$. $1.8 \text{ mg}$ water $(H_{2}O, M=18 \text{ g/mol})$: Moles = $1.8 \times 10^{-3} \text{ g} / 18 \text{ g/mol} = 10^{-4} \text{ mol}$. Each molecule has $3$ atoms $(2H + 1O)$. Total atoms = $10^{-4} \times 3 \times N_{A} = 3 \times 10^{-4} \times N_{A}$ $(III)$.
$B$. $9.8 \text{ mg}$ sulphuric acid $(H_{2}SO_{4}, M=98 \text{ g/mol})$: Moles = $9.8 \times 10^{-3} \text{ g} / 98 \text{ g/mol} = 10^{-4} \text{ mol}$. Each molecule has $7$ atoms $(2H + 1S + 4O)$. Total atoms = $10^{-4} \times 7 \times N_{A} = 7 \times 10^{-4} \times N_{A}$ $(IV)$.
$C$. $1.8 \text{ mg}$ carbon $(C, M=12 \text{ g/mol})$: Moles = $1.8 \times 10^{-3} \text{ g} / 12 \text{ g/mol} = 1.5 \times 10^{-4} \text{ mol}$. Total atoms = $1.5 \times 10^{-4} \times N_{A}$ $(II)$.
$D$. $5.85 \text{ mg}$ salt (NaCl,$M=58.5 \text{ g/mol}$): Moles = $5.85 \times 10^{-3} \text{ g} / 58.5 \text{ g/mol} = 10^{-4} \text{ mol}$. Each formula unit has $2$ atoms $(1Na + 1Cl)$. Total atoms = $10^{-4} \times 2 \times N_{A} = 2 \times 10^{-4} \times N_{A}$ $(I)$.
Thus,the correct match is $A-III, B-IV, C-II, D-I$.

(B) At $STP$,the molar volume of an ideal gas is $22.4 \ L/mol$.
Number of moles $(n)$ = $\frac{\text{Volume at STP}}{22.4 \ L/mol} = \frac{1.4187 \ L}{22.4 \ L/mol} \approx 0.0633 \ mol$.
Number of molecules = $\text{Number of moles} \times N_A$ (Avogadro constant).
Number of molecules = $0.0633 \times 6.022 \times 10^{23} \approx 3.812 \times 10^{22}$ molecules.

(B) The molar mass of urea $(CO(NH_{2})_{2})$ is $60 \ g \ mol^{-1}$.
Number of moles of urea = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.4 \ g}{60 \ g \ mol^{-1}} = 0.09 \ mol$.
Each molecule of urea contains $4$ hydrogen atoms.
Therefore,the number of moles of hydrogen atoms = $0.09 \ mol \times 4 = 0.36 \ mol$.
Number of hydrogen atoms = $\text{Number of moles} \times N_{A} = 0.36 \times 6.022 \times 10^{23} = 2.168 \times 10^{23}$ atoms.