A current of dry air was passed first through a series of bulbs containing a solution of C 6 H 5 – N O 2 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol. (T = 284 K) loss in weight of the solvent bulbs was 0.0685 g. Calculate the loss in weight of the solution bulbs.

Question

A current of dry air was passed first through a series of bulbs containing a solution of C 6 H 5 – N O 2 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol. (T = 284 K) loss in weight of the solvent bulbs was 0.0685 g. Calculate the loss in weight of the solution bulbs., 4.60 g, 5.20 g, 2.50 g, 2.05 g

MARKS App · Accepted Answer

If n 2 and n 1 are the number of moles of solute and solvent Then n 2 n 1 = mM 1000 Here m = molality M = molar mass of the solvent. n 2 n 1 = 0.725 × 46 1000 i . e . , 0.0334 Thus Weight loss in solution bulb ∝ P s Weight loss in solution bulb ∝ P o - P s Now , P o - P s P s = n 2 n 1 ⇒ 0.0685 P s = 0.0334 ∴ P s = 2.05 g

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