Answer:- Given: Boiling point elevation (\(\Delta T_b\)) = 1.76 K K_b of acetic acid = 3.07 K kg mol\(^{-1}\) Mass of acetic acid = 50 g To find: Molality of the solution Formula: \(\Delta T_b = K_b \cdot m\) Calculation: Using the formula and rearranging, we get, \[ m = \frac{\Delta T_b}{K_b} = \frac{1.76 \, \text{K}}{3.07 \, \text{K kg}^{-1}} = 0.573 \, \text{mol kg}^{-1} \] The molality of the solution is 0.573 m.