What is the relationship between the mole fraction of a solute (X_A) and its molality (m) if the molar mass of solvent is 100 (g/mol).

• a) X_A / 10(1 - X_A)
• b) X_A / 100(1 - X_A)
• c) 10(1 - X_A) / X_A
• d) 10(X_A) / (1 - X_A)

The relationship between the mole fraction of a solute (X_A) and its molality (m) can be derived using the definitions of these terms. Given that the molar mass of the solvent is 100 g/mol, we can express the molality in terms of mole fraction.

Definitions

• Mole Fraction (X_A): This is the ratio of the number of moles of solute to the total number of moles in the solution.
• Molality (m): This is defined as the number of moles of solute per kilogram of solvent.

Calculating Molality

To find the relationship, we start with the formula for molality:

m = (n_A / mass of solvent in kg)

Since the molar mass of the solvent is 100 g/mol, 1 kg of solvent corresponds to 10 moles of solvent.

Relating Mole Fraction and Molality

Let n_A be the number of moles of solute and n_B be the number of moles of solvent. The mole fraction of the solute is:

X_A = n_A / (n_A + n_B)

Substituting n_B = 10 (for 1 kg of solvent), we get:

X_A = n_A / (n_A + 10)

Finding the Correct Option

Rearranging gives:

n_A = 10X_A / (1 - X_A)

Now, substituting this into the molality formula, we find:

m = (10X_A / (1 - X_A)) / 1 = 10X_A / (1 - X_A)

Final Answer

The correct relationship is:

d) 10(X_A) / (1 - X_A)