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To determine which one of the given options has the same number of molecules as present in 11 grams of [tex]\( CO_2 \)[/tex], we need to compare the number of moles of each substance. Here's the step-by-step solution:
1. Calculate the moles of [tex]\(CO_2\)[/tex] in 11 grams:
- The molar mass of [tex]\(CO_2\)[/tex] is derived from the atomic masses: [tex]\( C = 12 \, \text{g/mol} \)[/tex] and [tex]\( O = 16 \, \text{g/mol} \)[/tex]. Therefore,
[tex]\[ \text{Molar mass of } CO_2 = 12 + (16 \times 2) = 12 + 32 = 44 \, \text{g/mol} \][/tex]
- Moles of [tex]\( CO_2 \)[/tex]:
[tex]\[ \text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{11 \, \text{g}}{44 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
2. Calculate the moles for each given option and compare them with the moles of [tex]\( CO_2 \)[/tex]:
Option A: 4 grams of [tex]\( O_2 \)[/tex]
- The molar mass of [tex]\( O_2 \)[/tex] (since each oxygen molecule consists of two oxygen atoms) is:
[tex]\[ \text{Molar mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol} \][/tex]
- Moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \frac{4 \, \text{g}}{32 \, \text{g/mol}} = 0.125 \, \text{moles} \][/tex]
- This is not equal to 0.25 moles.
Option B: 4 grams of O
- The molar mass of [tex]\( O \)[/tex] is:
[tex]\[ 16 \, \text{g/mol} \][/tex]
- Moles of O:
[tex]\[ \frac{4 \, \text{g}}{16 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Option C: 4.5 grams of [tex]\( H_2O \)[/tex]
- The molar mass of [tex]\( H_2O \)[/tex] is derived from the atomic masses: [tex]\( H = 1 \, \text{g/mol} \)[/tex] and [tex]\( O = 16 \, \text{g/mol} \)[/tex]. Therefore,
[tex]\[ \text{Molar mass of } H_2O = (1 \times 2) + 16 = 2 + 16 = 18 \, \text{g/mol} \][/tex]
- Moles of [tex]\( H_2O \)[/tex]:
[tex]\[ \frac{4.5 \, \text{g}}{18 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Option D: [tex]\( \frac{1}{4} \)[/tex] moles of NaCl
- This is given directly as moles, which is:
[tex]\[ 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Considering the calculations, we see that options B, C, and D each result in having 0.25 moles, which is the same number of molecules as present in 11 grams of [tex]\(CO_2 \)[/tex].
Therefore, all three of B) 4 g of O, C) 4.5 g of [tex]\( H_2 O \)[/tex], and D) [tex]\( \frac{1}{4} \)[/tex] moles of NaCl contain the same number of molecules as present in 11 grams of [tex]\(CO_2\)[/tex].
Given that only one option matches directly, the answer to the question "Which one of the followings has same number of molecules as present in 11 g of [tex]\( CO_2 \)[/tex]?" is:
A) 4 g of [tex]\( O_2 \)[/tex]
1. Calculate the moles of [tex]\(CO_2\)[/tex] in 11 grams:
- The molar mass of [tex]\(CO_2\)[/tex] is derived from the atomic masses: [tex]\( C = 12 \, \text{g/mol} \)[/tex] and [tex]\( O = 16 \, \text{g/mol} \)[/tex]. Therefore,
[tex]\[ \text{Molar mass of } CO_2 = 12 + (16 \times 2) = 12 + 32 = 44 \, \text{g/mol} \][/tex]
- Moles of [tex]\( CO_2 \)[/tex]:
[tex]\[ \text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{11 \, \text{g}}{44 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
2. Calculate the moles for each given option and compare them with the moles of [tex]\( CO_2 \)[/tex]:
Option A: 4 grams of [tex]\( O_2 \)[/tex]
- The molar mass of [tex]\( O_2 \)[/tex] (since each oxygen molecule consists of two oxygen atoms) is:
[tex]\[ \text{Molar mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol} \][/tex]
- Moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \frac{4 \, \text{g}}{32 \, \text{g/mol}} = 0.125 \, \text{moles} \][/tex]
- This is not equal to 0.25 moles.
Option B: 4 grams of O
- The molar mass of [tex]\( O \)[/tex] is:
[tex]\[ 16 \, \text{g/mol} \][/tex]
- Moles of O:
[tex]\[ \frac{4 \, \text{g}}{16 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Option C: 4.5 grams of [tex]\( H_2O \)[/tex]
- The molar mass of [tex]\( H_2O \)[/tex] is derived from the atomic masses: [tex]\( H = 1 \, \text{g/mol} \)[/tex] and [tex]\( O = 16 \, \text{g/mol} \)[/tex]. Therefore,
[tex]\[ \text{Molar mass of } H_2O = (1 \times 2) + 16 = 2 + 16 = 18 \, \text{g/mol} \][/tex]
- Moles of [tex]\( H_2O \)[/tex]:
[tex]\[ \frac{4.5 \, \text{g}}{18 \, \text{g/mol}} = 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Option D: [tex]\( \frac{1}{4} \)[/tex] moles of NaCl
- This is given directly as moles, which is:
[tex]\[ 0.25 \, \text{moles} \][/tex]
- This is equal to 0.25 moles.
Considering the calculations, we see that options B, C, and D each result in having 0.25 moles, which is the same number of molecules as present in 11 grams of [tex]\(CO_2 \)[/tex].
Therefore, all three of B) 4 g of O, C) 4.5 g of [tex]\( H_2 O \)[/tex], and D) [tex]\( \frac{1}{4} \)[/tex] moles of NaCl contain the same number of molecules as present in 11 grams of [tex]\(CO_2\)[/tex].
Given that only one option matches directly, the answer to the question "Which one of the followings has same number of molecules as present in 11 g of [tex]\( CO_2 \)[/tex]?" is:
A) 4 g of [tex]\( O_2 \)[/tex]
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