Skip to content
Exercise · Q35

Q.A solution containing 7.5 g7.5\ \text{g} of a non-volatile, non-electrolyte solute in 250 mL250\ \text{mL} of solution has an osmotic pressure of 0.6 atm0.6\ \text{atm} at 300 K300\ \text{K}. Calculate the molar mass of the solute. (R=0.0821 L atm mol−1K−1R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}.)

West Bengal WbchseTextbookSubjectiveImportance★★★★★
70% · 35/50 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 →

From π=CRT\pi=CRT, C=π/(RT)=0.6/(0.0821×300)=0.6/24.63≈0.02436 mol L−1C = \pi/(RT) = 0.6/(0.0821\times300) = 0.6/24.63 \approx 0.02436\ \text{mol L}^{-1}. Moles of solute in 250 mL=0.250 L250\ \text{mL} = 0.250\ \text{L}: n2=C×V≈0.02436×0.250≈0.006090 moln_2 = C\times V \approx 0.02436\times0.250 \approx 0.006090\ \text{mol}. Molar mass $M_2 = w_2/n_2 = 7.5/0.006090 …

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.