$273.15\,\, - \,\,271\,\, = \,\,{K_f}\, \times \,\,\frac{{5\,\, \times \,\,1000}}{{180\, \times \,\,100}}\,\,\,$
$ \Rightarrow \,\,2.15\,\, = \,\,\frac{{{K_f}\, \times \,\,5\, \times \,\,10}}{{342}}$ ........ $(1)$
${(\Delta {T_f})_{gulcose}}\, = \,\,{K_f}m\,\, = \,\,{K_f}\, \times \,\,\frac{{5\,\, \times \,\,1000}}{{180\,\, \times \,\,100}}$
$ \Rightarrow \,\,\,{(\Delta {T_f})_{glucose}}\,\, = \,\,{K_f}\, \times \,\,\frac{{5\,\, \times \,\,10}}{{180}}$ ......... $(2)$
${\text{(1)}}$ માંથી ${\text{(2)}}$ ને ભાગતા $\,\,\frac{{{\text{2}}{\text{.15}}}}{{{{{\text{(}}\Delta {{\text{T}}_{\text{f}}}{\text{)}}}_{{\text{glucose}}}}}}\,\, = \,\,\frac{{180}}{{342}}$
${(\Delta {T_f})_{glucose}}\,\, = \,\,4.085\,\,K$
${({T_f})_{glucose\,\,solution}}\,\, = \,\,{({T_f})_{water}}\, - \,\,{(\Delta {T_f})_{glucose}}$
$ = \,\,273.15\,\, - \,\,4.085\,\,\, = \,\,269.065\,\, = \,\,269.07\,\,K$
[ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]
$\mathrm{NaCl}$ નું સંપૂર્ણ વિયોજન થાય છે તે ધારી લો.
(અહીં : $25^o C$ પર બાષ્પ દબાણના મૂલ્યો અનુક્રમે બેન્ઝિન $= 12.8\, kPa,$ ટોલ્યુઇન $= 3.85 \,kPa$)
[આપેલ $K_b (H_2O) = 0.52\, K\, kg\, mol^{-1}]$