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Q.Define :

(a) Mole fraction
(b) Molality
(c) Molarity. Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass. OR Define and explain elevation in boiling point. How can you calculate the molecular mass of a non-volatile solute with it ?
Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2024Subjective· 5mImportance★★★★★
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Definitions of mole fraction, molality and molarity, then a worked calculation giving mole fraction of ethylene glycol ≈ 0.068 in a 20%-by-mass solution.

  1. Mole fraction (x): The ratio of the number of moles of one component of a solution to the total number of moles of all components (solute + solvent). For a two-component solution of A in B: x_A = n_A / (n_A + n_B). It is a dimensionless quantity, and the mole fractions of all components in a solution add up to 1.
  2. Molality (m): The number of moles of solute dissolved per kilogram (1000 g) of solvent (not solution). m = (moles of solute) / (mass of solvent in kg). Unlike molarity, molality does not change with temperature, since it is based on mass, not volume.
  3. Molarity (M): The number of moles of solute dissolved per litre (dm³) of SOLUTION (not solvent). M = (moles of solute) / (volume of solution in L). Calculation — mole fraction of ethylene glycol (C₂H₆O₂) in a 20% (by mass) solution: Take 100 g of solution: mass of ethylene glycol = 20 g, mass of water (solvent) = 80 g. Molar mass of C₂H₆O₂ = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 g/mol. Moles of ethylene glycol, n₁ = 20 / 62 = 0.3226 mol. Molar mass of water = 18 g/mol; moles of water, n₂ = 80 / 18 = 4.444 mol. Mole fraction of ethylene glycol, x₁ = n₁ / (n₁ + n₂) = 0.3226 / (0.3226 + 4.444) = 0.3226 / 4.767 ≈ 0.068 …

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