Question

which of the following actions increases the concentration of an aqueous solution? increasing the solute (tea) and solvent (water) in equal amounts; increasing the amount of the solute (tea); decreasing the amount of the solute (tea); increasing the amount of the solvent (water)

Explanation:

To determine which action increases the concentration of an aqueous solution, we recall that concentration (for a solution with solute and solvent) is generally defined as the amount of solute relative to the amount of solvent (or solution). The formula for concentration (e.g., mass/volume or moles/volume) is \( \text{Concentration} = \frac{\text{Amount of Solute}}{\text{Amount of Solvent (or Solution)}} \).

1. Analyzing "increasing the solute (tea) and solvent (water) in equal amounts": If we increase both solute and solvent by the same amount, the ratio of solute to solvent (or solution) remains approximately the same. For example, if we have 1g of solute in 10g of solvent (concentration \( \frac{1}{10} = 0.1 \)), and we add 1g of solute and 1g of solvent, we now have 2g of solute in 11g of solution, and \( \frac{2}{11} \approx 0.18 \)? Wait, no, wait. Wait, if it's a dilute solution, maybe the change is small, but actually, if we consider the ratio \( \frac{Solute}{Solvent} \), initially \( \frac{S}{V} \), after adding \( \Delta S \) and \( \Delta V \) where \( \Delta S=\Delta V \), the new ratio is \( \frac{S + \Delta S}{V+\Delta V} \). Let's take numbers: \( S = 1 \), \( V = 10 \), \( \Delta S=\Delta V = 1 \). New ratio: \( \frac{2}{11}\approx0.18 \), original was \( 0.1 \). Wait, that's an increase? But maybe in the context of a solution where the solute is dissolved, maybe the initial solution is already at a certain concentration, and adding equal amounts might not be the same as increasing concentration. Wait, maybe I made a mistake here. Alternatively, if we think of concentration as \( \frac{\text{Solute}}{\text{Solution}} \), initial \( \frac{1}{11} \), after adding 1 solute and 1 solvent, \( \frac{2}{12}=\frac{1}{6}\approx0.166 \), which is less than \( 0.18 \) but more than \( 0.1 \). Hmm, maybe this option is not the best.

1. Analyzing "increasing the amount of the solute (tea)": If we increase the amount of solute while keeping the amount of solvent constant, the ratio \( \frac{\text{Solute}}{\text{Solvent}} \) (or \( \frac{\text{Solute}}{\text{Solution}} \)) will increase. For example, if we have 1g of solute in 10g of solvent (concentration \( \frac{1}{10} = 0.1 \) or \( \frac{1}{11}\approx0.09 \) for solution), and we add 1g of solute, now we have 2g of solute in 10g of solvent (concentration \( \frac{2}{10} = 0.2 \) or \( \frac{2}{11}\approx0.18 \)), which is an increase.

1. Analyzing "decreasing the amount of the solute (tea)": Decreasing the solute while keeping solvent constant will decrease the concentration, so this is incorrect.

1. Analyzing "increasing the amount of the solvent (water)": Increasing the solvent while keeping solute constant will decrease the concentration (since the solute is spread out more), so this is incorrect.

So the action that increases the concentration is increasing the amount of the solute (tea).

Answer:

B. increasing the amount of the solute (tea)