What is the molality of a solution made by dissolving 137.9g of sucrose in 414.1g of water?

Answer: 2.71 moles of solute for every 1 kg of solvent.

Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent. This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water. 500. mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample. 115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent. In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water. 103g water⋅1.353 moles NaNO3500. g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg-1≈2.71 mol kg-1 The answer is rounded to three sig figs.