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JEE Advance - Chemistry (2014 - Paper 1 Offline - No. 3)

A compound H2X with molar weight of 80g is dissolved in a solvent having density of 0.4 gml–1 . Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is
Answer
8

Explanation

Understanding the Concepts

  • Molarity (M): Moles of solute per liter of solution.

  • Molality (m): Moles of solute per kilogram of solvent.

  • Density: Mass per unit volume.

    • Steps to Calculate Molality

    1. Find the Mass of Solute (H₂X):

    • Molarity = moles of solute / volume of solution (in liters)

    • 3.2 M solution means 3.2 moles of H₂X are present in 1 liter of solution.

    • Mass of H₂X = moles $$ \times $$ molar mass = 3.2 moles $$ \times $$ 80 g/mole = 256 g

    1. Find the Mass of Solvent:

    • Assuming no change in volume, the volume of the solution remains 1 liter.

    • Density = mass / volume

    • Mass of solvent = density $$ \times $$ volume = 0.4 g/mL $$ \times $$ 1000 mL = 400 g

    1. Convert Mass of Solvent to Kilograms:

    • 1 kg = 1000 g

    • Mass of solvent = 400 g $$ \times $$ (1 kg / 1000 g) = 0.4 kg

    1. Calculate Molality:

    • Molality = moles of solute / mass of solvent (in kg)

    • Molality = 3.2 moles / 0.4 kg = 8 mol/kg

    Answer:

    The molality of the 3.2 molar solution is 8 mol/kg.

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