2.0 molal aqueous solution of an electrolyte `X_(2) Y_(3)` is 75% ionised. The boiling point of the solution a 1 atm is `( K_(b(H_(2)O)) = 0.52K` kg `mol^(-1)`)

Options

A. 2.74 .76 K
B. 377 K
C. 376.4K
D. 377.16 K

Answer

To solve the problem step by step, we will follow the outlined process to find the boiling point of the solution.

### Step 1: Understand the Problem
We have a 2.0 molal aqueous solution of the electrolyte \(X_2Y_3\) which is 75% ionized. We need to find the boiling point of this solution. The boiling point elevation constant for water, \(K_b(H_2O)\), is given as \(0.52 \, K \, kg \, mol^{-1}\).

### Step 2: Calculate the Van't Hoff Factor (i)
The formula for the Van't Hoff factor \(i\) is given by:
\[ i = 1 + (n - 1) \cdot \alpha \]
where:
- \(n\) is the number of particles the solute dissociates into,
- \(\alpha\) is the degree of ionization.

For the electrolyte \(X_2Y_3\):
- It dissociates into \(2X^+ + 3Y^{3-}\), which gives a total of \(5\) ions.
- Thus, \(n = 5\).
- The degree of ionization \(\alpha = 0.75\) (75%).

Substituting these values into the equation:
\[ i = 1 + (5 - 1) \cdot 0.75 \]
\[ i = 1 + 4 \cdot 0.75 = 1 + 3 = 4 \]

### Step 3: Calculate the Boiling Point Elevation (\(\Delta T_b\))
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(K_b = 0.52 \, K \, kg \, mol^{-1}\),
- \(m = 2.0 \, molal\).

Substituting the values:
\[ \Delta T_b = 4 \cdot 0.52 \cdot 2.0 \]
\[ \Delta T_b = 4 \cdot 1.04 = 4.16 \, K \]

### Step 4: Calculate the New Boiling Point (\(T_b\))
The boiling point of pure water is \(100^\circ C\) or \(373 \, K\). The new boiling point of the solution is given by:
\[ T_b = T_{b, \text{water}} + \Delta T_b \]
\[ T_b = 373 \, K + 4.16 \, K = 377.16 \, K \]

### Final Answer
The boiling point of the solution is \(377.16 \, K\).

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