An aqueous solution freezes at –2.55 0 C. What is its boiling point ( K_ b of water =0.52…
Chemistry · JEE Advanced · NTA Exams — Solutions
An aqueous solution freezes at –2.550C. What is its boiling point (\(K_{b}\) of water \(=0.52 \mathrm{~K} / \mathrm{m} ;\) \(K_{f}\) of water
\(=1.86 \mathrm{~K} / \mathrm{m}) ?\)
\(=1.86 \mathrm{~K} / \mathrm{m}) ?\)
- 107.00C
- 100.60C
- 100.10C
- 100.70C
Answer
(D) 100.7 0 C
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