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Answer :
- Calculate the molar mass of $NH_3$: $M_{NH_3} = 14.01 + 3 \times 1.008 = 17.034$ g/mol.
- Calculate the number of moles of $NH_3$: $n = \frac{6.3 \times 10^{24}}{6.022 \times 10^{23}} = 10.46$.
- Calculate the mass of the $NH_3$ sample: $mass = 10.46 \times 17.034 = 178.20$ grams.
- The closest answer from the options is 182 grams, so the final answer is $\boxed{182 \text{ grams}}$.
### Explanation
1. Problem Analysis
We are given a sample of $NH_3$ (ammonia) that contains $6.3
\times 10^{24}$ molecules. Our goal is to find the mass of this sample in grams. To do this, we'll use the molar mass of $NH_3$ and Avogadro's number.
2. Calculating Molar Mass of Ammonia
First, we need to calculate the molar mass of $NH_3$. The molar mass of nitrogen (N) is approximately 14.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.008 g/mol. Since there are three hydrogen atoms in $NH_3$, the molar mass of $NH_3$ is:
$$M_{NH_3} = M_N + 3 \times M_H = 14.01 + 3 \times 1.008 = 14.01 + 3.024 = 17.034 \frac{g}{mol}$$
So, the molar mass of $NH_3$ is approximately 17.034 g/mol.
3. Calculating Number of Moles
Next, we need to find the number of moles of $NH_3$ in the sample. We can use Avogadro's number, which is approximately $6.022 \times 10^{23}$ molecules/mol. The number of moles (n) is given by:
$$n = \frac{N}{N_A} = \frac{6.3 \times 10^{24}}{6.022 \times 10^{23}} = 10.46164065094653$$
So, there are approximately 10.46 moles of $NH_3$ in the sample.
4. Calculating Mass of the Sample
Now, we can calculate the mass of the $NH_3$ sample by multiplying the number of moles by the molar mass:
$$mass = n \times M_{NH_3} = 10.46164065094653 \times 17.034 = 178.20358684822318$$
So, the mass of the $NH_3$ sample is approximately 178.20 grams.
5. Final Answer
Comparing our calculated mass (approximately 178.20 grams) with the given options:
a) 157 grams
b) 198 grams
c) 182 grams
d) 198 grams
The closest option to our calculated mass is 182 grams. Therefore, the correct answer is:
c) 182 grams
### Examples
Understanding molar mass and Avogadro's number is crucial in various real-world applications. For instance, in chemical reactions, knowing the exact mass of reactants is essential for predicting the yield of products. In environmental science, it helps in quantifying pollutants in air or water samples. Moreover, in pharmaceuticals, precise mass calculations ensure accurate drug dosages, which is vital for patient safety.
- Calculate the number of moles of $NH_3$: $n = \frac{6.3 \times 10^{24}}{6.022 \times 10^{23}} = 10.46$.
- Calculate the mass of the $NH_3$ sample: $mass = 10.46 \times 17.034 = 178.20$ grams.
- The closest answer from the options is 182 grams, so the final answer is $\boxed{182 \text{ grams}}$.
### Explanation
1. Problem Analysis
We are given a sample of $NH_3$ (ammonia) that contains $6.3
\times 10^{24}$ molecules. Our goal is to find the mass of this sample in grams. To do this, we'll use the molar mass of $NH_3$ and Avogadro's number.
2. Calculating Molar Mass of Ammonia
First, we need to calculate the molar mass of $NH_3$. The molar mass of nitrogen (N) is approximately 14.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.008 g/mol. Since there are three hydrogen atoms in $NH_3$, the molar mass of $NH_3$ is:
$$M_{NH_3} = M_N + 3 \times M_H = 14.01 + 3 \times 1.008 = 14.01 + 3.024 = 17.034 \frac{g}{mol}$$
So, the molar mass of $NH_3$ is approximately 17.034 g/mol.
3. Calculating Number of Moles
Next, we need to find the number of moles of $NH_3$ in the sample. We can use Avogadro's number, which is approximately $6.022 \times 10^{23}$ molecules/mol. The number of moles (n) is given by:
$$n = \frac{N}{N_A} = \frac{6.3 \times 10^{24}}{6.022 \times 10^{23}} = 10.46164065094653$$
So, there are approximately 10.46 moles of $NH_3$ in the sample.
4. Calculating Mass of the Sample
Now, we can calculate the mass of the $NH_3$ sample by multiplying the number of moles by the molar mass:
$$mass = n \times M_{NH_3} = 10.46164065094653 \times 17.034 = 178.20358684822318$$
So, the mass of the $NH_3$ sample is approximately 178.20 grams.
5. Final Answer
Comparing our calculated mass (approximately 178.20 grams) with the given options:
a) 157 grams
b) 198 grams
c) 182 grams
d) 198 grams
The closest option to our calculated mass is 182 grams. Therefore, the correct answer is:
c) 182 grams
### Examples
Understanding molar mass and Avogadro's number is crucial in various real-world applications. For instance, in chemical reactions, knowing the exact mass of reactants is essential for predicting the yield of products. In environmental science, it helps in quantifying pollutants in air or water samples. Moreover, in pharmaceuticals, precise mass calculations ensure accurate drug dosages, which is vital for patient safety.
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