The problem is asking to calculate the atomic mass of an element $X$ that exists in two isotopic forms:

• X-6 with a mass of $6.015 \, \text{amu}$
• X-7 with a mass of $7.016 \, \text{amu}$ and a natural abundance of 92.5%.

Weighted Average Formula:

The atomic mass $M_{\text{avg}}$ is given by the formula:

$M_{\text{avg}} = (f_{\text{X-6}} \times m_{\text{X-6}}) + (f_{\text{X-7}} \times m_{\text{X-7}})$

Where:

• $f_{\text{X-6}}$ and $f_{\text{X-7}}$ are the fractional abundances of X-6 and X-7, respectively.
• $m_{\text{X-6}} = 6.015 \, \text{amu}$
• $m_{\text{X-7}} = 7.016 \, \text{amu}$
• $f_{\text{X-7}} = 92.5\% = 0.925$
• Since the total abundance must sum to 1, $f_{\text{X-6}} = 1 - f_{\text{X-7}} = 1 - 0.925 = 0.075$

Calculation:

Now, substitute the values into the weighted average formula:

$M_{\text{avg}} = (0.075 \times 6.015) + (0.925 \times 7.016)$

$M_{\text{avg}} = (0.451125) + (6.4898)$

$M_{\text{avg}} = 6.940925 \, \text{amu}$

So, the atomic mass of element $X$ is approximately $6.94 \, \text{amu}$.

Final Answer:

The correct choice is (d) 6.94 amu.

Would you like further details or have any questions?

Here are five questions related to this problem:

1. What is the difference between atomic mass and isotopic mass?
2. How does the natural abundance of isotopes affect the average atomic mass?
3. Why do some elements have multiple naturally occurring isotopes?
4. Can you describe other methods for calculating atomic mass from isotopic data?
5. How is atomic mass used in practical chemistry applications?

Tip: Always ensure that the sum of isotopic abundances equals 100% or 1 when performing weighted average calculations.