x mole of $\mathrm{KCl}$and y mole of ${\mathrm{BaCl}}_2$ are both dissolved in 1 kg of water. Given that$\mathrm{x} + \mathrm{y} = 0.1$ and $\mathrm{Kr}$ for water is $1.85 \mathrm{k}/\mathrm{molal}$, what is the observed range of $\mathrm{\Delta }{T}_f$, if the ratio of x to y is varied? 

$\mathrm{x}$mole of $\mathrm{KCl}$and $\mathrm{y}$ mole of ${\mathrm{BaCl}}_2$ 

$\begin{array}{l}\mathrm{KCl}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}⟹{\mathrm{k}}^{+}+{\mathrm{Cl}}^{(-)}\\ i=2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x\mathit{ }\mathrm{mole})\end{array}$     $\begin{array}{c}{\mathrm{BaCl}}_2\rightleftharpoons {\mathrm{Ba}}^{2+}+2{\mathrm{a}}^{(-)}\\ i=3\cdot \text{ (y mole) }\end{array}$

$\begin{array}{l}x+y=0.1\text{ solvent }=1\mathrm{Kg}.\\ \mathrm{\Delta }{T}_f=i{k}_{t}m\\ =(2\times x+3\times y)\times 1.85\end{array}$

$1.85\times 2\times 0.1<\mathrm{\Delta }{T}_f<1.85\times 3\times 0.1$

or $0.37<\mathrm{\Delta }{T}_f<0.555$

Thus, the ratio of x to y is varied 0.37° to 0.555°.