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If we assume that the density of the solution is almost equal to that of pure water (1 kg/L), then we find that 1 L of solution will weigh 1 kg. If 1 kg of solution contains 0.0001 M of sucrose , then it is 0.0001 m in sucrose, so the molarity and molality are almost the same. I think they are numerically the same to several decimal places.
[13.9]. Thus, if you form a solution by mixing 0.200 mol of NaOH (40.0 g) and 0.500 kg of water (500 g), the concentration of the solution is (0.200 mol)/(0.500 kg) = 0.400 m (that is, 0.400 molal) in NaOH. Notice the difference between the definitions of molarity and molality.