Let's determine the number of molecules in each of the given scenarios and then compare them.
### a. 0.25 mole of oxygen gas (O₂)
First, recall that 1 mole of any substance contains Avogadro's number of molecules, which is approximately [tex]\( 6.022 \times 10^{23} \)[/tex].
- Number of molecules in 0.25 mole of oxygen gas:
[tex]\[ \text{Molecules}_{O_2} = 0.25 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Molecules}_{O_2} = 1.5055 \times 10^{23} \][/tex]
### b. 20 grams of NaOH
To find the number of molecules, we first need to calculate the number of moles of NaOH. The molar mass of NaOH can be calculated as follows:
- Atomic mass of Na (Sodium) = 23
- Atomic mass of O (Oxygen) = 16
- Atomic mass of H (Hydrogen) = 1
Therefore, the molar mass of NaOH = 23 + 16 + 1 = 40 grams/mol.
- Number of moles of NaOH:
[tex]\[ \text{Moles}_{NaOH} = \frac{20 \, \text{grams}}{40 \, \text{grams/mol}} \][/tex]
[tex]\[ \text{Moles}_{NaOH} = 0.5 \][/tex]
- Number of molecules in 0.5 mole of NaOH:
[tex]\[ \text{Molecules}_{NaOH} = 0.5 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Molecules}_{NaOH} = 3.011 \times 10^{23} \][/tex]
### c. 12.4 litres of ammonia (NH₃) at NTP
At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 litres.
- Number of moles of NH₃:
[tex]\[ \text{Moles}_{NH_3} = \frac{12.4 \, \text{litres}}{22.4 \, \text{litres/mol}} \][/tex]
[tex]\[ \text{Moles}_{NH_3} \approx 0.5536 \][/tex]
- Number of molecules in 0.5536 mole of NH₃:
[tex]\[ \text{Molecules}_{NH_3} = 0.5536 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Molecules}_{NH_3} \approx 3.333 \times 10^{23} \][/tex]
### d. 1.9 grams of hydrogen gas (H₂)
To find the number of molecules, we first need to calculate the number of moles of H₂. The molar mass of H₂ can be calculated as follows:
- Atomic mass of H (Hydrogen) = 1
- Since H₂ is diatomic, molar mass = 2 \times 1 = 2 grams/mol.
- Number of moles of H₂:
[tex]\[ \text{Moles}_{H_2} = \frac{1.9 \, \text{grams}}{2 \, \text{grams/mol}} \][/tex]
[tex]\[ \text{Moles}_{H_2} = 0.95 \][/tex]
- Number of molecules in 0.95 mole of H₂:
[tex]\[ \text{Molecules}_{H_2} = 0.95 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{Molecules}_{H_2} = 5.7209 \times 10^{23} \][/tex]
### Conclusion
Now compare the number of molecules in each case:
- a. Oxygen gas: [tex]\( 1.5055 \times 10^{23} \)[/tex]
- b. NaOH: [tex]\( 3.011 \times 10^{23} \)[/tex]
- c. Ammonia: [tex]\( 3.333 \times 10^{23} \)[/tex]
- d. Hydrogen gas: [tex]\( 5.7209 \times 10^{23} \)[/tex]
Thus, option (d) - 1.9 grams of hydrogen gas has the greatest number of molecules.
