QJEEAdv14Ch47f5: JEE Advanced 2014 Chemistry - Class 11 JEEAdv14Ch47f5 Solution: A compound H 2 X with molar weight of 80g is dissolved in a…

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

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