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Question 42 of 50

Q.(i) What is van't Hoff factor?

(ii) 36 g of glucose dissolved per litre of the solution has an osmotic pressure 4.98 bar at 300 K. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its molar concentration? (Given, molar mass of glucose = 180 g mol-1) [1+2]
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 3mImportance★★★★★
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Using the first data point to find R, then applying π=CRT\pi = CRT again with the new osmotic pressure gives C ≈ 0.061 mol/L.

  1. van't Hoff factor: i=observed (experimental) colligative propertycalculated (theoretical/normal) colligative property=normal molar massobserved/abnormal molar massi = \dfrac{\text{observed (experimental) colligative property}}{\text{calculated (theoretical/normal) colligative property}} = \dfrac{\text{normal molar mass}}{\text{observed/abnormal molar mass}} It measures the deviation of a solute's actual colligative behaviour from ideal (non-dissociating, non-associating) behaviour: i>1i > 1 for dissociating solutes (electrolytes), i<1i < 1 for associating solutes, and i=1i = 1 for an ideal non-electrolyte like glucose.
  2. Finding the new molar concentration: Given: 36 g glucose per litre, π1=4.98\pi_1 = 4.98 bar at T=300T = 300 K, molar mass of glucose =180= 180 g mol−1^{-1}. Concentration: C1=36180=0.2C_1 = \dfrac{36}{180} = 0.2 mol L−1^{-1}. Since glucose is a non-electrolyte (i=1i=1), π=CRT\pi = CRT: …

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