Chemistry Q141 – AP-EAMCET 2026

The mole fraction of

C H_{3} O H

in an aqueous solution is 0.02. What is the molality of this solution?

Clue 1 of 3: Understanding Mole Fraction and Molality

Mole fraction relates the moles of a component to the total moles in the solution, while molality relates the moles of solute to the mass of the solvent (in kg).

🧠 Full Solution Path

Step 1: Determine Moles of Solute and Solvent✦ Active

Given the mole fraction of $C H_{3} O H$ ( $X_{C H_{3} O H}$ ) is $0.02$ . In an aqueous solution, water ( $H_{2} O$ ) is the solvent. The sum of mole fractions of all components in a solution is 1.

X_{C H_{3} O H} + X_{H_{2} O} = 1

So, the mole fraction of water ( $X_{H_{2} O}$ ) is:

X_{H_{2} O} = 1 - X_{C H_{3} O H} = 1 - 0.02 = 0.98

To easily work with these values, assume a basis of 1 mole of total solution. This means:

Moles of C H_{3} O H (n_{C H_{3} O H}) = 0.02 mol

Moles of H_{2} O (n_{H_{2} O}) = 0.98 mol

💡 Teacher's Secret Hint

Assuming a basis for calculation is a common and effective strategy in stoichiometry problems involving mole fractions.

Step 2: Calculate the Mass of the Solvent in Kilograms○ Expand

The solvent is water. To calculate molality, we need the mass of the solvent in kilograms. The molar mass of water ( $M_{H_{2} O}$ ) is approximately $18.015 g/mol$ (or $18 g/mol$ ). For calculations, we'll use $18 g/mol$ .

Mass of H_{2} O = n_{H_{2} O} \times M_{H_{2} O}

Mass of H_{2} O = 0.98 mol \times 18 g/mol = 17.64 g

Convert the mass from grams to kilograms:

Mass of H_{2} O (kg) = \frac{17.64 g}{1000 g/kg} = 0.01764 kg

💡 Teacher's Secret Hint

Always double-check units! Molality requires the mass of solvent in kilograms. Forgetting to convert grams to kilograms is a common mistake.

Step 3: Calculate the Molality of the Solution○ Expand

Now, we have the moles of solute ( $C H_{3} O H$ ) and the mass of the solvent ( $H_{2} O$ ) in kilograms. We can calculate the molality ( $m$ ) using its definition:

Molality (m) = \frac{Moles of solute}{Mass of solvent (kg)}

m = \frac{0.02 mol C H_{3} O H}{0.01764 kg H_{2} O}

m \approx 1.13378 mol/kg

Rounding to two decimal places, the molality is $1.13 m$ .

💡 Teacher's Secret Hint

Pay attention to significant figures and rounding instructions in multiple-choice questions. In this case, comparing to options, two decimal places are appropriate.

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Chemistry Question 141 – AP-EAMCET 2026

Clue 1 of 3: Understanding Mole Fraction and Molality

🧠 Full Solution Path