The relationship between mole fraction of the solute and molality and of its solution in Ammonia would beAns: 58.82(Xa)÷1-Xa=mExplanation please

The relationship between the mole fraction of a solute and its molality in a solution, particularly in ammonia, can be understood through the definitions of these terms and how they interact with each other. Let's break this down step by step.

Understanding Mole Fraction and Molality

Mole fraction (Xa) is defined as the ratio of the number of moles of a component to the total number of moles of all components in the solution. For a solute, it can be expressed as:

• X_a = n_solute / (n_solute + n_solvent)

Where n_solute is the number of moles of the solute and n_solvent is the number of moles of the solvent.

On the other hand, molality (m) is defined as the number of moles of solute per kilogram of solvent:

• m = n_solute / mass_solvent (kg)

Connecting the Two Concepts

To establish a relationship between mole fraction and molality, we can manipulate the definitions. Let's denote:

• n_solute = number of moles of solute
• n_solvent = number of moles of solvent (ammonia in this case)
• mass_solvent = mass of solvent in kilograms

From the mole fraction formula, we can rearrange it to express n_solvent in terms of Xa:

• n_solvent = (n_solute / X_a) - n_solute

Now, substituting this into the molality formula gives us a way to express molality in terms of mole fraction:

Deriving the Relationship

We know that:

• mass_solvent = n_solvent × M_solvent

Where M_solvent is the molar mass of the solvent (for ammonia, it is approximately 17 g/mol). Thus, we can express molality as:

• m = n_solute / (n_solvent × M_solvent / 1000)

Substituting for n_solvent gives:

• m = n_solute / (((n_solute / X_a) - n_solute) × M_solvent / 1000)

This equation can be simplified to show how molality relates to mole fraction. The specific relationship you mentioned, 58.82(X_a) ÷ (1 - X_a) = m, indicates that for a given mole fraction, you can calculate the corresponding molality.

Practical Example

Let’s say you have a solution where the mole fraction of the solute (ammonia) is 0.5. Plugging this into the equation:

• m = 58.82(0.5) ÷ (1 - 0.5)
• m = 29.41

This means that at a mole fraction of 0.5, the molality of the solution is approximately 29.41 mol/kg. This relationship is particularly useful in chemistry for calculating concentrations in various solutions, especially when dealing with solvents like ammonia.

In summary, the relationship between mole fraction and molality in ammonia solutions can be derived through their definitions and manipulated algebraically to yield a clear formula that connects the two. Understanding this relationship is crucial for solving various problems in chemistry related to solutions.