If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?(a) 21.3 g/mole(b) 23.1 g/mole(c) 32.1 g/mole(d) 1.23 g/moleI have been asked this question at a job interview.I need to ask this question from Colligative Properties and Determination of Molar Mass in section Solutions of Chemistry – Class 12

If the relative lowering of pressure of o-xylene is 0.005 due to addit

1.	If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?(a) 21.3 g/mole(b) 23.1 g/mole(c) 32.1 g/mole(d) 1.23 g/moleI have been asked this question at a job interview.I need to ask this question from Colligative Properties and Determination of Molar Mass in section Solutions of Chemistry – Class 12
Answer» The CORRECT option is (a) 21.3 g/mole Best explanation: GIVEN, ∆p1/p^01 = 0.005 Mass of solute, mW = 0.5 grams Mass of solvent, mS = 500 grams Number of moles of solute, NS = 0.5/MW, where MW is the molecular weight of the solute. Number of moles of solvent, nSolvent = 500 grams/(106 g/mole) = 4.7 mole Since mS< Δp1/p^01 = N2/(n1 + n2) ≈ n2/n1 On substituting values, 0.005 = (0.5/MW)/4.7 On SOLVING, MW = 21.3 g/mole.

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