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Question

The Van't Hoff factor for 0.1 M Ba(NO₃)₂ solution is 2.74. The degree of dissociation is:

A

91.30%

B

87%

C

100%

D

74%

Step-by-Step Solution

The relationship between the Van't Hoff factor (ii), the degree of dissociation (α\alpha), and the number of ions produced per formula unit (nn) is given by the formula: i=1+(n−1)αi = 1 + (n - 1)\alpha

  1. Identify nn: Barium nitrate dissociates as: Ba(NO3)2→Ba2++2NO3−Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-. Total ions (nn) = 1+2=31 + 2 = 3.

  2. Substitute values: Given i=2.74i = 2.74. 2.74=1+(3−1)α2.74 = 1 + (3 - 1)\alpha

  • 2.74=1+2α2.74 = 1 + 2\alpha
  1. Solve for α\alpha: 2α=2.74−12\alpha = 2.74 - 1 2α=1.742\alpha = 1.74
  • α=1.742=0.87\alpha = \frac{1.74}{2} = 0.87
  1. Convert to percentage:
  • Percentage degree of dissociation = 0.87×100%=87%0.87 \times 100\% = 87\%.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSolutionsfactorsolutiondegreedissociation

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