Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (K$$_b$$/K$$_f$$) is

We are given elevation in boiling point for 1.5 molal glucose solution: $$\Delta T_b = 4$$ K and depression in freezing point for 4.5 molal glucose solution: $$\Delta T_f = 4$$ K.

Using the boiling point elevation formula $$\Delta T_b = K_b \times m$$ gives $$4 = K_b \times 1.5$$ and hence $$K_b = \frac{4}{1.5} = \frac{8}{3}$$ K kg mol$$^{-1}$$. Similarly, the freezing point depression formula $$\Delta T_f = K_f \times m$$ yields $$4 = K_f \times 4.5$$ so $$K_f = \frac{4}{4.5} = \frac{8}{9}$$ K kg mol$$^{-1}$$. The ratio is then $$\frac{K_b}{K_f} = \frac{8/3}{8/9} = \frac{8}{3} \times \frac{9}{8} = \frac{9}{3} = 3$$.

Therefore, the ratio $$K_b/K_f = \textbf{3}$$.