`0.250g` of an element M, reacts with excess fluorine to produce `0.547g` of the hexafluoride `MF_(6)`. What is the elemejnt : `[Cr = 52, Mo = 96, S = 32, Te = 127.6]` .

Options

A. Cr
B. MO
C. S
D. Te

Explanation

To solve the problem, we need to determine the molar mass of the element M that reacts with fluorine to form the hexafluoride MF₆. We can use the information given in the question to find the molar mass of the element M.

### Step-by-Step Solution:

1. **Write the reaction equation:**
The reaction of element M with fluorine can be represented as:
\[ M + 6F \rightarrow MF_6 \]

2. **Identify the mass of M and MF₆:**
We are given:
- Mass of element M = 0.250 g
- Mass of hexafluoride MF₆ = 0.547 g

3. **Calculate the mass of fluorine in MF₆:**
The molar mass of fluorine (F) is 19 g/mol. Since there are 6 fluorine atoms in MF₆, the total mass of fluorine in one mole of MF₆ is:
\[ \text{Mass of } 6F = 6 \times 19 = 114 \text{ g/mol} \]

4. **Determine the molar mass of MF₆:**
The molar mass of MF₆ can be expressed as:
\[ \text{Molar mass of } MF_6 = \text{Molar mass of M} + \text{Mass of } 6F = X + 114 \]
where \(X\) is the molar mass of element M.

5. **Set up the equation using the mass of MF₆:**
Using the mass of MF₆ produced, we can set up the equation:
\[ \frac{0.250}{X} = \frac{0.547}{X + 114} \]

6. **Cross-multiply to solve for X:**
Cross-multiplying gives:
\[ 0.250 \times (X + 114) = 0.547 \times X \]
Expanding this:
\[ 0.250X + 28.5 = 0.547X \]

7. **Rearranging the equation:**
Rearranging the equation to isolate \(X\):
\[ 28.5 = 0.547X - 0.250X \]
\[ 28.5 = 0.297X \]

8. **Solve for X:**
Dividing both sides by 0.297:
\[ X = \frac{28.5}{0.297} \approx 95.95 \text{ g/mol} \]

9. **Identify the element:**
Now we compare the calculated molar mass of M (95.95 g/mol) with the given options:
- Cr = 52 g/mol
- Mo = 96 g/mol
- S = 32 g/mol
- Te = 127.6 g/mol

The closest match is Molybdenum (Mo), which has a molar mass of approximately 96 g/mol.

### Final Answer:
The element M is **Molybdenum (Mo)**.