Respuesta :
Explanation:
Relation between [tex]k_{f}[/tex], molality and temperature is as follows.
         T = [tex]K_{f} \times m \times i[/tex]
It is also known as depression between freezing point where, i is the Van't Hoff factor.
Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.
     i for [tex]CaCl_{2}[/tex] = 3
      i for glucose = 1
      i for NaCl = 2
Depression in freezing point will have a negative sign. Therefore, d
depression in freezing point for the given species is as follows.
    [tex]T_{CaCl_{2}} = -1.86 \times 0.45 \times 3[/tex]
         = [tex]-2.511^{o}C[/tex]
    [tex]T_{glucose} = 1.86 \times 0.45 \times 1[/tex]
          = [tex]-0.837^{o}C[/tex]
    [tex]T_{NaCl} = -1.86 \times 0.45 \times 2[/tex]
          = [tex]-1.674^{o}C[/tex]
Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.
           Glucose < NaCl < [tex]CaCl_{2}[/tex]
