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 formula of the compound.

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

### Step 2: Calculate the simplest ratio of the elements
To find the empirical formula, we need to divide the number of moles of each element by the smallest number of moles among them.

- The smallest number of moles is for Cu: 0.539 mol.

Now, we divide each mole value by 0.539:

- For Na:
\[ \frac{1.08}{0.539} \approx 2 \]

- For Cu:
\[ \frac{0.539}{0.539} = 1 \]

- For F:
\[ \frac{2.16}{0.539} \approx 4 \]

### Step 3: Write the empirical formula
From the ratios obtained, we can write the empirical formula as:
\[ \text{Na}_2\text{CuF}_4 \]

### Step 4: Analyze the osmotic pressure information
We know that the osmotic pressure of the compound is three times that of urea at the same molar concentration.

The formula for osmotic pressure (\(\pi\)) is given by:
\[ \pi = iCRT \]
Where:
- \(i\) = Van't Hoff factor
- \(C\) = concentration
- \(R\) = universal gas constant
- \(T\) = temperature

For urea (a non-electrolyte), the Van't Hoff factor \(i\) is 1. Since the osmotic pressure of the compound is three times that of urea, we can conclude:
\[ i = 3 \]

### Step 5: Determine the dissociation of the compound
The compound \(\text{Na}_2\text{CuF}_4\) dissociates in solution as follows:
\[ \text{Na}_2\text{CuF}_4 \rightarrow 2\text{Na}^+ + \text{Cu}^{2+} + 4\text{F}^- \]
This gives us a total of:
- 2 from Na
- 1 from Cu
- 4 from F

Thus, the total number of particles (ions) produced is:
\[ 2 + 1 + 4 = 7 \]

However, we need to check if this matches the Van't Hoff factor of 3. The correct dissociation should yield a total of 3 particles.

### Step 6: Correct the dissociation
Upon re-evaluating, we realize that the compound dissociates as:
\[ \text{Na}_2\text{CuF}_4 \rightarrow 2\text{Na}^+ + \text{CuF}_4^{2-} \]
This gives:
- 2 from Na
- 1 from CuF4

Thus, the total number of particles is:
\[ 2 + 1 = 3 \]
This matches the Van't Hoff factor of 3.

### Conclusion
The empirical formula of the compound is confirmed as:
\[ \text{Na}_2\text{CuF}_4 \]

### Final Answer
The formula of the compound is \(\text{Na}_2\text{CuF}_4\).