Q.What is the molality of an aqueous solution of KBr having freezing point –3.72°C (K$_f$ for water is 1.86 K kg mol$^{-1}$)?

$\Delta T_f=K_fm=1.86\times0.1=0.186$; freezing point $=0-0.186=-0.186°C$.

Depression in freezing point: $\Delta T_f = K_f \times m = 1.86 \times 0.1 = 0.186\ K$

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$\Delta T_f=iK_fm$ with $i=2$ for KBr: $m=3.72/(2\times1.86)=1$ molal.

KBr is a strong electrolyte, dissociating completely into 2 ions ($K^+ + Br^-$), so the van't Hoff factor $i=2$ must be used:

$\Delta T_f = i \times K_f \times m$

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Since $\Delta T_f = K_f\times m$ (molality in $mol\,kg^{-1}$), the SI unit of $K_f$ works out to $K\,kg\,mol^{-1}$.

The cryoscopic (molal freezing-point-depression) constant is defined by $\Delta T_f = K_f \times m$, where $\Delta T_f$ is in kelvin and molality $m$ is in $mol\,kg^{-1}$. Rearranging, $K_f = \Delta T_f/m$, so …