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MHT CET Chemistry Solutions 2026 MHT CET 2026 (15 April Shift 1)

MHT CET Chemistry Question (2026) — Solution

Question

5 g urea is dissolved in 100 g water. Find the amount of glucose to be dissolved in 120 g of water so that the boiling points of both solutions will be the same. [Molar mass of urea = 60 g mol ^ -1 & Molar mass of glucose = 180 g mol ^ -1 ]

Options

  1. A. 17 g
  2. B. 19 g
  3. C. 18 g
  4. D. 20 g

Answer

C. 18 g

Step-by-step solution

Elevation in boiling point is given by T_b = K_b m For the boiling points to be the same, the molality of both solutions must be equal. Molality of urea solution = W_ urea M_ urea W_ water (kg) = 5 60 0.1 = 5 6 m Let the mass of glucose be x g. Molality of glucose solution = W_ glucose M_ glucose W_ water (kg) = x 180 0.12 = x 21.6 m Equating the molalities: 5 6 = x 21.6 x = 5 21.6 6 = 5 3.6 = 18 g Answer: 18 g

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