Final answer:
Under typical conditions, the 5.5% m/v NaOH solution (B) would be hypertonic compared to the 0.25 M NaOH. Therefore, the correct answer would be option B:
Step-by-step explanation:
Given a 0.25 M NaOH solution (A) and a 5.5% m/v NaOH solution (B), we can deduce that one of these solutions will have a higher concentration of solutes, making it hypertonic, while the other will have a lower concentration, making it hypotonic.
The percentage concentration (5.5% m/v) means that there are 5.5 grams of NaOH in every 100 milliliters of solution. To compare it with the molar concentration, we would have to know the density of the solution and convert the percentage to molarity. However, without this information, a direct comparison is not possible.
Assuming typical conditions where the density of NaOH solution is high enough that 5.5% m/v represents a higher solute concentration than 0.25 M, solution B would be hypertonic and solution A would be hypotonic. Water molecules will move from the hypotonic solution (A) to the hypertonic solution (B) until both solutions reach an isotonic state.