Finding Molality When Mole Fraction is Given
In the Solutions chapter, converting between different concentration terms is a guaranteed question type in JEE and NEET. One of the most frequently asked conversions is finding the Molality (m) when only the Mole Fraction (X) is provided.
Many textbooks give you massive, complicated formulas to memorize for this. But what if you forget the formula during the exam? Let's learn a purely logical, foolproof trick to solve this in under 60 seconds.
Video Tutorial: The "Assume 1 Mole" Trick
Watch Abhishek Sengar sir from CHEMCA demonstrate how to use simple interpretations to unlock the hidden mass of the solvent.
Step-by-Step Problem Breakdown
Problem Statement:
In an aqueous solution of Sulfuric Acid (H2SO4), the mole fraction of H2SO4 is given as 0.1. Calculate the Molality of the solution.
If the mole fraction of a solute is 0.1, the easiest way to solve the problem is to assume the total moles of the entire solution is exactly 1 Mole.
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Find Moles of Solute and Solvent:
Let Total Moles of Solution = 1 mole.
Therefore, Moles of Solute (H2SO4) = 0.1 moles.
Since MolesSolute + MolesSolvent = Total Moles,
Moles of Solvent (H2O) = 1 - 0.1 = 0.9 moles. -
Find the Mass of the Solvent:
Molality requires the mass of the solvent in kg (or grams in the numerator). We must convert our 0.9 moles of water into grams.
Molar mass of H2O = 18 g/mol.
Mass of H2O = 0.9 moles × 18 g/mol = 16.2 grams. -
Apply the Molality Formula:
m = (Moles of Solute × 1000) / (Mass of Solvent in grams)m = (0.1 × 1000) / 16.2
m = 100 / 16.2 = 1000 / 162 mol/kg. (You can simplify this fraction further depending on the MCQ options).
Fig: Notice how avoiding complex fraction formulas prevents calculation errors.
Practice Questions for JEE & NEET
Test your speed and conceptual depth with these two essential exam variations.
Question 1: The mole fraction of Glucose (C6H12O6) in an aqueous solution is 0.2. Calculate the Molality of the solution. (Molar mass of water = 18 g/mol).
Answer: 13.88 mol/kg (or 1000/72)
Step-by-Step Solution:
- Step 1: Assume 1 total mole.
Moles of Glucose (Solute) = 0.2 mol.
Moles of Water (Solvent) = 1 - 0.2 = 0.8 mol. - Step 2: Find mass of solvent in grams.
Mass of Water = 0.8 mol × 18 g/mol = 14.4 g. - Step 3: Apply Molality formula.
m = (0.2 × 1000) / 14.4
m = 200 / 14.4 = 2000 / 144 = 13.88 m.
Question 2 (Conceptual Theory): Between Molarity and Molality, which concentration term is preferred by scientists when performing experiments at varying temperatures, and why?
Answer: Molality is preferred because it is Independent of Temperature.
Reasoning:
Molarity (M) is defined as Moles per Liter of Volume. Because the volume of a liquid expands or contracts with changes in temperature, the Molarity of a solution will change as the temperature changes!
Molality (m), however, is defined as Moles per Kilogram of Mass. Since mass does not change with temperature, Molality remains perfectly constant regardless of thermal conditions. (Mole Fraction is also independent of temperature for the same reason).
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