the vapour pressure of solution containing 2 gram of NaCl in hundred gram of water at hundred degree centigrade is 753. 2 mm of HG then the degree of dissociation of NaCl is what

To determine the degree of dissociation of NaCl in your solution, we first need to understand a few key concepts related to colligative properties and vapor pressure. The degree of dissociation refers to the fraction of the solute that separates into ions when dissolved in a solvent. In the case of sodium chloride (NaCl), it dissociates into sodium ions (Na⁺) and chloride ions (Cl^-).

Understanding Vapor Pressure and Colligative Properties

When a non-volatile solute like NaCl is added to a solvent such as water, the vapor pressure of the solution decreases compared to that of the pure solvent. This decrease can be quantified using Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

Given Data

• Mass of NaCl = 2 g
• Mass of water = 100 g
• Vapor pressure of the solution = 753.2 mmHg
• Vapor pressure of pure water at 100°C = 760 mmHg

Calculating the Moles of NaCl

First, we need to calculate the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol. Thus, the moles of NaCl can be calculated as follows:

Moles of NaCl = Mass / Molar Mass

Moles of NaCl = 2 g / 58.44 g/mol ≈ 0.0342 moles

Calculating the Moles of Water

Next, we calculate the moles of water. The molar mass of water (H₂O) is about 18.02 g/mol:

Moles of Water = Mass / Molar Mass

Moles of Water = 100 g / 18.02 g/mol ≈ 5.55 moles

Finding the Mole Fraction of Water

The total number of moles in the solution is the sum of the moles of NaCl and the moles of water:

Total Moles = Moles of NaCl + Moles of Water

Total Moles = 0.0342 + 5.55 ≈ 5.5842 moles

The mole fraction of water (X_H2O) is then calculated as:

X_H2O = Moles of Water / Total Moles

X_H2O = 5.55 / 5.5842 ≈ 0.9948

Applying Raoult's Law

According to Raoult's Law, the vapor pressure of the solution (P_solution) can be expressed as:

P_solution = X_H2O × P_H2O

Substituting the known values:

P_solution = 0.9948 × 760 mmHg ≈ 756.4 mmHg

Calculating the Degree of Dissociation

The observed vapor pressure of the solution is 753.2 mmHg, which is slightly less than the calculated value of 756.4 mmHg. The difference indicates that some of the NaCl has dissociated into ions. The degree of dissociation (α) can be calculated using the formula:

ΔP = P_H2O - P_solution

Where ΔP is the change in vapor pressure. Thus:

ΔP = 760 mmHg - 753.2 mmHg = 6.8 mmHg

Now, we can relate this change to the degree of dissociation:

ΔP = α × P_H2O × i × X_H2O

Here, i is the van 't Hoff factor for NaCl, which is 2 (since it dissociates into two ions). Rearranging gives:

α = ΔP / (P_H2O × i × X_H2O)

Substituting the known values:

α = 6.8 mmHg / (760 mmHg × 2 × 0.9948) ≈ 0.0045

Final Thoughts

This means that the degree of dissociation of NaCl in your solution is approximately 0.0045, or 0.45%. This low value suggests that the dissociation of NaCl is not complete, which can happen under certain conditions, such as high ionic strength or specific interactions in the solution.