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Manipal MET2012ChemistrySolutions

The elevation in boiling point would be highest for

Options

  1. A0.08 M BaCl ₂
  2. B0.15 M KCl
  3. C0.10 M Glucose
  4. D0.06 M Ca ( NO ₃ )₂

Correct answer

B. 0.15 M KCl

Step-by-step solution

array ll & T_b= 1000 K_b w m W i & T_b i w m W 1000 ( K_b= constant ) array or T_b i M aligned ( Molality & = w m W 1000 and assuming, & Molarity, M= molality, m) aligned Now, for the given solutions, (a) For 0.08 M BaCl ₂, i M=3 0.08=0.24 (b) For 0.15 M KCl , i M=2 0.15=0.30 (c) For 0.10 M glucose, i M=1 0.10=0.10 (d) For 0.06 MCa ( NO ₃ )₂, i M=3 0.06=0.18 Hence, T_b is maximum for 0.15 M KCl solution.

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