An element X of atomic mass 25.0 exists as `X)_(4)` in benzene to the extent of `100%`. When `10.30g` of saturated solution of X in benzene is added to 20.0 g of benzene, the depression in freezing point of the resulting solution is 0.51 K. If `K_(f)` for benzene is `"5.1 K kg mol"^(-1)`, the solubility of X in 100 g of benzene will be

Options

A. 3.0 g
B. 2.7 g
C. 0.30 g
D. 0.27 g

Explanation

To solve the problem, we need to determine the solubility of element X in 100 g of benzene based on the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- Element X has an atomic mass of 25.0 g/mol and exists as \(X_4\).
- The mass of the saturated solution of X in benzene is 10.30 g.
- The mass of benzene added is 20.0 g.
- The depression in freezing point (\(\Delta T_f\)) is 0.51 K.
- The freezing point depression constant (\(K_f\)) for benzene is 5.1 K kg/mol.

### Step 2: Calculate the Total Mass of Benzene
The total mass of benzene in the solution after adding the 20.0 g of pure benzene to the saturated solution is:
\[ \text{Total mass of benzene} = 20.0 \, \text{g} + (10.30 \, \text{g} - w) \]
where \(w\) is the mass of solute \(X_4\) in the saturated solution.

### Step 3: Use the Freezing Point Depression Formula
The formula for freezing point depression is:
\[ \Delta T_f = \frac{K_f \cdot m}{1000} \]
where \(m\) is the molality of the solution, defined as:
\[ m = \frac{n}{\text{mass of solvent in kg}} \]
Here, \(n\) is the number of moles of solute.

### Step 4: Calculate the Moles of Solute
The molar mass of \(X_4\) is:
\[ \text{Molar mass of } X_4 = 4 \times 25.0 \, \text{g/mol} = 100.0 \, \text{g/mol} \]
Thus, the number of moles of solute is:
\[ n = \frac{w}{100.0} \]

### Step 5: Set Up the Equation
Substituting into the freezing point depression formula:
\[ 0.51 = \frac{5.1 \cdot \left(\frac{w}{100.0}\right)}{(30.30 - w)/1000} \]
This can be rearranged to:
\[ 0.51 = \frac{5.1 \cdot w \cdot 1000}{100.0 \cdot (30.30 - w)} \]

### Step 6: Solve for \(w\)
Cross-multiplying gives:
\[ 0.51 \cdot 100.0 \cdot (30.30 - w) = 5.1 \cdot w \cdot 1000 \]
Expanding and rearranging:
\[ 51.0 \cdot 30.30 - 51.0w = 5100w \]
\[ 1547.3 = 5151w \]
\[ w = \frac{1547.3}{5151} \approx 0.3 \, \text{g} \]

### Step 7: Calculate the Solubility in 100 g of Benzene
Now, we know that in the saturated solution (10.30 g), there are approximately 0.3 g of solute \(X_4\). The mass of benzene in the saturated solution is:
\[ 10.30 - 0.3 = 10.0 \, \text{g} \]
Thus, the solubility of \(X\) in 100 g of benzene can be calculated as:
\[ \text{Solubility} = \frac{0.3 \, \text{g}}{10.0 \, \text{g}} \times 100 \, \text{g} = 3.0 \, \text{g} \]

### Final Answer
The solubility of element X in 100 g of benzene is **3.0 g**.

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