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Q.The boiling point of benzene is 353.2 K. When 1.8 g of a non-volatile solute was dissolved in 90 g benzene the boiling point was raised to 354.1 K. Calculate the molecular mass of the solute. (Kb of benzene = 2.53 K kg mol-1)

Odisha ChseOdisha CHSE +2 Science Board Exam 2020Subjective· 3mImportance★★★★★
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Use elevation in boiling point, ΔTb=1000 Kb w2M2 w1\Delta T_b = \dfrac{1000\,K_b\,w_2}{M_2\,w_1}, and solve for the molar mass M2M_2.

Given:

  • Boiling point of pure benzene, Tb0=353.2T_b^0 = 353.2 K
  • Boiling point of solution, Tb=354.1T_b = 354.1 K
  • ΔTb=354.1−353.2=0.9\Delta T_b = 354.1 - 353.2 = 0.9 K
  • Mass of solute, w2=1.8w_2 = 1.8 g
  • Mass of solvent (benzene), w1=90w_1 = 90 g
  • KbK_b (benzene) =2.53= 2.53 K kg mol−1^{-1}

The elevation-in-boiling-point formula (with masses in grams):

ΔTb=1000×Kb×w2M2×w1\Delta T_b = \dfrac{1000 \times K_b \times w_2}{M_2 \times w_1}

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