What mass of a 1.23 m sodium carbonate (105.99 g/mol) solution will contain 0.23 mol of solute?
A)211 g
B)187 g
C)24.0 g
D)323 g
E)347 g

Question

Quizplus · Accepted Answer

To find the mass of the solution that contains 0.23 mol of sodium carbonate, first calculate the mass of the sodium carbonate alone by multiplying the number of moles by the molar mass: $0.23 \, \text{mol} \times 105.99 \, \text{g/mol} = 24.377 \, \text{g}$. Since the solution is 1.23 m (which means 1.23 mol of solute per kilogram of solvent), to find the mass of the solution that contains 0.23 mol of solute, we use the proportion: $1.23 \, \text{mol} \rightarrow 1000 \, \text{g} \, \text{of solution}$ and $0.23 \, \text{mol} \rightarrow x \, \text{g} \, \text{of solution}$. Solving for $x$ gives approximately $187 \, \text{g}$ of solution. However, this calculation mistake led to the wrong choice being selected. The correct approach involves directly calculating the total mass of the solution based on the molarity and the amount of solute. Given the molarity (molality, in fact, but the question seems to confuse molarity and molality) and the moles of solute, the correct calculation should directly lead to the mass of the solution needed to contain 0.23 mol of sodium carbonate without needing to calculate the mass of the solute separately or use the proportion method described. The correct calculation involves understanding that the molarity (or molality) and the amount of solute directly give us the necessary information to find the total mass of the solution, but the explanation provided mistakenly led to an incorrect choice. The correct mass should be calculated based on the given molarity and the amount of solute, taking into account the density or the specific way molarity (or molality) relates to the mass of the solution and the solute.

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