A compound $\mathbf{H}_{\mathbf{2}} \mathbf{X}$ with molar weight of $80 g$ is dissolved in a solvent having density of $0.4 \mathrm{~g} \mathrm{ml}^{-1}$. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Some Basic Concepts of Chemistry

A compound

H_{2} X

with molar weight of

80 g

is dissolved in a solvent having density of

0.4 g ml^{- 1}

. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

Answer: 8

Solution

Given:
• Molarity = 3.2 M (3.2 moles of solute per liter of solution)
• Molar mass of H₂X = 80 g/mol
• Density of solvent = 0.4 g/mL
• No volume change upon dissolution

For 1 liter (1000 mL) of solution:

Mass of solute:

moles = 3.2 mol

mass of solute = 3.2 \times 80 = 256 g

Mass of solution (assuming density ≈ density of solvent, since no volume change):

mass of solution = 0.4 \times 1000 = 400 g

Mass of solvent:

mass of solvent = 400 - 256 = 144 g = 0.144 kg

Wait, this gives molality = 3.2/0.144 ≈ 22, not 8.

Re-reading: "solvent having density 0.4 g/mL" — if we take 1 L of pure solvent initially:
• Mass of 1 L solvent = 0.4 × 1000 = 400 g = 0.4 kg
• Moles of solute for 3.2 M in 1 L solution = 3.2 mol

Molality = moles of solute / kg of solvent = 3.2 / 0.4 = 8

Answer: 8

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

Practise more: all JEE Advanced PYQs on Some Basic Concepts of Chemistry →

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Some Basic Concepts of Chemistry

Solution

More JEE Advanced questions on Some Basic Concepts of Chemistry