Skip to content
Question of 131

Q.If the relative lowering of vapour pressure of a dilute aqueous solution of a non-volatile solute is 0.0125, determine the molality of the solution.

Tripura TbseHigher Secondary (+2 Stage) Examination 2023Subjective· 2mImportance★★★★★
0% · 0/131 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For a dilute solution, the relative lowering of vapour pressure equals the mole fraction of solute; from this we can back-calculate the molality using the fact that 1000 g (1 kg) of water contains 55.55 mol.

Raoult's law (for a dilute solution of a non-volatile solute) gives the relative lowering of vapour pressure as equal to the mole fraction of the solute:

(p0 - p) / p0 = x2 = n2 / (n1 + n2)

Given: (p0-p)/p0 = 0.0125

For a dilute solution, n2 (moles of solute) is much smaller than n1 (moles of solvent, water), so:

x2 (approx) = n2 / n1

Number of moles of water in 1000 g (1 kg): n1 = 1000 g / 18 g/mol = 55.56 mol

So: n2 = x2 x n1 = 0.0125 x 55.56 = 0.694 mol

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.