# The freezing point of a $0.05$ molal solution of a nonelectrolyte in water is

$\begin{array}{1 1} -1.86^{\circ} \\ -0.93^{\circ} C \\ -0.093^{\circ}C \\ 0.93^{\circ}\;C \end{array}$

Answer :$-0.093^{\circ}C$
$- \Delta T_f =K_tm =(1.86\;K\;kg\;mol^{-1})(0.05 \;mol\;kg^{-1}) =0.093K.$
Freezing point $=-0.093 ^{\circ} C$