When `60.26g` of `VCl_(4)` was added to 1000g of solvent `"CC"l_(4)`, the freezing point of `"CC"l_(4)` was depresed by `5.415^(@)C`. `K_(f)` for `"CC"l_(4)` is `29.9`. Compare the number of moles of particles with the number predicted by the formula. Calculate the number of dimers, `V_(2)Cl_(8)` present.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of VCl₄
First, we need to determine the molecular weight of VCl₄. The molecular weight can be calculated as follows:

- Vanadium (V) = 51 g/mol
- Chlorine (Cl) = 35.5 g/mol

The molecular weight of VCl₄ = 51 + 4 × 35.5 = 51 + 142 = 193 g/mol.

Now, we can calculate the number of moles of VCl₄ in 60.26 g:

\[ \text{Moles of VCl₄} = \frac{\text{mass}}{\text{molecular weight}} = \frac{60.26 \, \text{g}}{193 \, \text{g/mol}} \approx 0.312 \, \text{mol} \]

### Step 2: Use the freezing point depression formula
The freezing point depression (\(ΔT_f\)) is given by the formula:

\[ ΔT_f = i \cdot K_f \cdot m \]

Where:
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_f\) = freezing point depression constant
- \(m\) = molality of the solution

Given:
- \(ΔT_f = 5.415 \, °C\)
- \(K_f = 29.9 \, °C \cdot kg/mol\)
- Mass of solvent (CCl₄) = 1000 g = 1 kg

First, we need to calculate the molality (m):

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.312 \, \text{mol}}{1 \, \text{kg}} = 0.312 \, \text{mol/kg} \]

Now, substituting into the freezing point depression formula:

\[ 5.415 = i \cdot 29.9 \cdot 0.312 \]

### Step 3: Solve for \(i\)
Rearranging the equation to solve for \(i\):

\[ i = \frac{5.415}{29.9 \cdot 0.312} \approx 0.56 \]

### Step 4: Determine the number of dimers
The van 't Hoff factor \(i\) indicates how many particles the solute dissociates into. For VCl₄, it can either remain as a molecule or form dimers (V₂Cl₈).

If \(x\) is the degree of dimerization, then:

\[ i = 1 - x + 2x = 1 + x \]

Setting \(i = 0.56\):

\[ 0.56 = 1 + x \implies x = 0.56 - 1 = -0.44 \]

Since \(x\) cannot be negative, we need to adjust our understanding. The correct interpretation is that if \(i\) is less than 1, it indicates that some of the VCl₄ is dimerizing.

Let’s denote the number of moles of VCl₄ that dimerize as \(y\). The total moles of particles would be:

\[ \text{Total moles} = (0.312 - y) + 2y = 0.312 + y \]

Setting this equal to the calculated \(i\):

\[ 0.312 + y = 0.56 \implies y = 0.56 - 0.312 = 0.248 \]

### Step 5: Calculate the number of dimers
Since each dimer (V₂Cl₈) consists of 2 moles of VCl₄, the number of dimers formed is:

\[ \text{Number of dimers} = \frac{y}{2} = \frac{0.248}{2} = 0.124 \]

### Final Answer
The number of dimers \(V_2Cl_8\) present is approximately **0.124 moles**.