The formula for depression of freezing point is:
[tex]\Delta T_f = k_fm[/tex] -(1)
where, [tex]\Delta T_f[/tex] is depression of freezing point, [tex]k_f[/tex] is freezing point depression constant and [tex]m[/tex] is molality.
Freezing point of pure water = [tex]T_{o}^{f} = 0.00^{o} C[/tex] (given)
Freezing point of glucose = [tex]T_{f} = -4.15^{o} C[/tex] (given)
Depression of freezing point, [tex]\Delta T_f[/tex] = [tex]T_{o}^{f} - T_f[/tex]
Substituting the values:
[tex]\Delta T_f = 0.00^{o} C - (-4.15^{o} C) = 4.15^{o} C[/tex]
Substituting the values in formula (1):
[tex]4.15^{o} C = 1.86^{o}C/m\times m[/tex]
[tex]m = \frac{4.15^{o} C}{1.86^{o}C/m} = 2.231 m[/tex]
Hence, the the molal concentration of glucose in the given solution is [tex]2.231 m[/tex].