Q.Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelctrolyte and its molar mass.

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

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

Since freezing point is depressed (lowered) relative to pure water's $0°C$:

$T_f(\text{solution}) = 0 - 0.186 = -0.186°C$

$\Delta T_f=iK_fm$ 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:

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

Given $\Delta T_f = 3.72°C$ (freezing point $-3.72°C$, i.e. depressed by $3.72°C$ from $0°C$), $K_f = 1.86\ K\,kg\,mol^{-1}$, $i=2$:

$m = \dfrac{\Delta T_f}{i\,K_f} = \dfrac{3.72}{2 \times 1.86} = \dfrac{3.72}{3.72} = 1\ mol\,kg^{-1}$

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 its SI unit is $K\,kg\,mol^{-1}$ (often quoted as $K\,kg\,mol^{-1}$ or equivalently $^\circ C\,kg\,mol^{-1}$).