Final answer:
The molality of an aqueous solution where the mole fraction of NaOH is 0.1 can be calculated by considering the mole fraction of NaOH and water in the solution. Assuming 0.1 moles of NaOH and 0.9 moles of water, the molality is found to be 6.17 mol/kg.
Step-by-step explanation:
The question is asking for the molality of an aqueous solution of NaOH with a mole fraction of 0.1. Molality is defined as the number of moles of solute per kilogram of solvent. To determine the molality given a mole fraction, we need to calculate the amount of NaOH and water in the solution.
First, we recall that the mole fraction (X) is defined as the number of moles of solute (nsolute) divided by the total number of moles in the solution (nsolute + nsolvent). Since we have XNaOH = 0.1, we can express this as: XNaOH = nNaOH / (nNaOH + nH2O). If we let nNaOH = 0.1 moles, to have a simple mole fraction of 0.1, we find that nH2O must be 0.9 moles because 0.1 / (0.1 + 0.9) = 0.1.
Pure water has a molar mass of approximately 18.02 g/mol, therefore 0.9 moles of water weigh approximately 16.22 grams (0.9 moles * 18.02 g/mol). As we are interested in the molality, which requires kilograms of solvent, this converts to 0.01622 kg. Hence, the molality of the NaOH solution would be 0.1 moles / 0.01622 kg = 6.17 mol/kg.