A portein has been isolated as sodium salt with their molecular formula `Na_(x)P` (this notation means that `xNa^(+)` ions are associated with a negatively charged protein `P^(-x))`. A solution of this salt was prepared by dissolving `0.25g` of this sodium slat of protein in `10g` of water adn ebulliscopic analysis revealed that soluion boils at temperature `5.93 xx 10^(-3).^(@)C` higher than the normal boiling point of pure water. `K_(b)` of water 0`.52 kg mol^(-1)`. Also elemental analysis revealed that the salt contain `1%` sodium metal by weight. Deduce molecular formula and determine molecular weight of acidic form of protein `H_(X)P`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a sodium salt of a protein with the formula `Na_xP`. The solution of this salt shows a boiling point elevation, which we can use to find the molality and ultimately the molecular weight of the acidic form of the protein `H_XP`.

### Step 2: Use the Boiling Point Elevation Formula
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(\Delta T_b\) = boiling point elevation
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 3: Calculate the Van 't Hoff Factor \(i\)
Since the sodium salt dissociates into \(x\) sodium ions and one negatively charged protein ion, the total number of particles is:
\[ i = x + 1 \]

### Step 4: Calculate the Molality \(m\)
The molality \(m\) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Given that we have 0.25 g of the sodium salt and 10 g of water, we convert the mass of water to kg:
\[ \text{mass of solvent} = 10 \text{ g} = 0.01 \text{ kg} \]
Now, let \(M\) be the molar mass of the sodium salt. The number of moles of solute is:
\[ \text{moles of solute} = \frac{0.25 \text{ g}}{M} \]
Thus, the molality \(m\) becomes:
\[ m = \frac{0.25/M}{0.01} = \frac{25}{M} \]

### Step 5: Substitute Values into the Boiling Point Elevation Equation
From the problem, we have:
\[ \Delta T_b = 5.93 \times 10^{-3} \, ^\circ C \]
And \(K_b\) for water is given as \(0.52 \, \text{kg mol}^{-1}\). Substituting these values into the boiling point elevation formula:
\[ 5.93 \times 10^{-3} = (x + 1)(0.52) \left(\frac{25}{M}\right) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ M = (x + 1)(0.52)(25) / (5.93 \times 10^{-3}) \]

### Step 7: Analyze Sodium Content
The problem states that the sodium salt contains 1% sodium by weight. Therefore, we can set up the equation:
\[ \frac{x \cdot 23}{0.25} = 0.01 \implies x \cdot 23 = 0.0025 \implies x = \frac{0.0025}{23} \approx 0.0001087 \]
This means:
\[ x = 20 \]

### Step 8: Substitute \(x\) Back to Find \(M\)
Now substituting \(x = 20\) back into the equation for \(M\):
\[ M = (20 + 1)(0.52)(25) / (5.93 \times 10^{-3}) = 21 \cdot 0.52 \cdot 25 / (5.93 \times 10^{-3}) \]
Calculating this gives:
\[ M \approx 4556 \, \text{g/mol} \]

### Step 9: Determine the Molecular Formula
The molecular formula of the acidic form of the protein is:
\[ H_{20}P \]

### Final Answer
The molecular weight of the acidic form of the protein \(H_XP\) is approximately **4556 g/mol**.

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