A compound contains 1.08 mol of Na , 0.539 mol of cu and 2.16 mol of F it's aqueous solution shows osmotic pressure which is three times that of urea having same molar concentration. The formula of the compound is :

Options

A. `Na_4[CuF_6]`
B. `Na[CuF_4]`
C. `Na_2[CuF_4]`
D. `Na_2[CuF_3]`

Answer

To solve the problem step by step, we will analyze the information given and derive the molecular formula of the compound.

### Step 1: Identify the moles of each element
The compound contains:
- Sodium (Na): 1.08 mol
- Copper (Cu): 0.539 mol
- Fluorine (F): 2.16 mol

### Step 2: Determine the simplest mole ratio
To find the simplest ratio of the elements, we divide the number of moles of each element by the smallest number of moles among them, which is 0.539 mol (for Cu).

- Na: \( \frac{1.08}{0.539} \approx 2 \)
- Cu: \( \frac{0.539}{0.539} = 1 \)
- F: \( \frac{2.16}{0.539} \approx 4 \)

Thus, the simplest mole ratio is approximately:
- Na: 2
- Cu: 1
- F: 4

### Step 3: Write the empirical formula
Based on the simplest mole ratio, the empirical formula of the compound is:
\[ \text{Na}_2\text{CuF}_4 \]

### Step 4: Understand the osmotic pressure relationship
The problem states that the osmotic pressure of the solution is three times that of urea at the same molar concentration.

- For urea, the van 't Hoff factor (i) is 1, as it does not dissociate in solution.
- For the compound, the van 't Hoff factor (i) is 3, indicating that the compound dissociates into three particles in solution.

### Step 5: Confirm the dissociation of the compound
The dissociation of Na2CuF4 can be represented as:
\[ \text{Na}_2\text{CuF}_4 \rightarrow 2\text{Na}^+ + \text{Cu}^{2+} + 4\text{F}^- \]

This dissociation results in a total of 3 ions (2 Na+ and 1 Cu2+) and 4 F- ions, confirming that the total number of particles is consistent with the van 't Hoff factor of 3.

### Conclusion
Thus, the formula of the compound is:
\[ \text{Na}_2\text{CuF}_4 \]

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