The specific heat at constant volume for the monatomic argon is `0.075 kcal//kg-K`, whereas its gram molecular specific heat is `C_(V) = 2.98 cal//mol//K`. The mass of the argon atom is (Avogadro's number `= 6.02 xx 10^(23) "molecules"//"mol")`

Options

A. `6.60 xx 10^(-23) g`
B. `3.30 xx 10^(-23) g`
C. `2.20 xx 10^(-23) g`
D. `13.20 xx 10^(-23) g`

Explanation

To find the mass of an argon atom given its specific heat at constant volume and the molar specific heat, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Specific heat at constant volume for argon (C_V) = 0.075 kcal/kg-K
- Molar specific heat (C_V) = 2.98 cal/mol-K
- Avogadro's number (N_A) = \(6.02 \times 10^{23}\) molecules/mol

2. **Convert Units:**
- Convert the specific heat from kcal to cal:
\[ 0.075 \text{ kcal/kg-K} = 0.075 \times 1000 \text{ cal/kg-K} = 75 \text{ cal/kg-K} \]

3. **Use the Relationship Between Molar Specific Heat and Molecular Weight:**
- The relationship is given by:
\[ C_V = \text{Molecular Weight} \times \text{Specific Heat} \]
- Rearranging gives:
\[ \text{Molecular Weight} = \frac{C_V (\text{molar})}{C_V (\text{specific})} \]

4. **Substitute the Values:**
- Substitute the values into the equation:
\[ \text{Molecular Weight} = \frac{2.98 \text{ cal/mol-K}}{75 \text{ cal/kg-K}} \]

5. **Calculate the Molecular Weight:**
- Performing the division:
\[ \text{Molecular Weight} = \frac{2.98}{75} \approx 0.03973 \text{ kg/mol} = 39.73 \text{ g/mol} \]

6. **Find the Mass of One Argon Atom:**
- The mass of one atom can be found using Avogadro's number:
\[ \text{Mass of one atom} = \frac{\text{Molecular Weight}}{N_A} \]
- Substitute the values:
\[ \text{Mass of one atom} = \frac{39.73 \text{ g/mol}}{6.02 \times 10^{23} \text{ molecules/mol}} \]

7. **Calculate the Mass of One Argon Atom:**
- Performing the division:
\[ \text{Mass of one atom} \approx \frac{39.73}{6.02 \times 10^{23}} \approx 6.60 \times 10^{-23} \text{ g} \]

### Final Answer:
The mass of an argon atom is approximately \(6.60 \times 10^{-23}\) grams.

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