What is the molality of an aqueous solution of KBr having freezing point −3.72ºC (Kf for water is 1.86 K kg mol−1)?
Q.2 (v) What is the molality of an aqueous solution of KBr having freezing point −3.72ºC (Kf for water is 1.86 K kg mol−1)?
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Answer: 1 mol kg−1
\(\Delta T_f = 0 - (-3.72) = 3.72\) K. KBr dissociates into K+ and Br−, so \(i = 2\). \(\Delta T_f = i K_f m \Rightarrow m = \dfrac{3.72}{2 \times 1.86} = 1\) mol kg\(^{-1}\). (Without the van't Hoff factor the value would be 2 m, which is the total molality of ions.)
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· 28 Oct 2026
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