(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) A certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

In this scenario, some of the 0.10 M solution of the salt is spilled. As a result, the volume of the solution decreases. However, when the solution is spilled, both the solute and the solvent are spilled, and therefore the moles of solute relative to the new volume remain constant. In other words, the concentration of the solution left in the container remains unchanged.

In this case, we have a 0.10 M aqueous solution of salt, and some water evaporates over time. As water evaporates, the volume of the solvent decreases, but the amount of solute (salt) remains constant. The moles of solute (n) do not change, but the volume (V) decreases. Consequently, using the formula \( C=\frac{n}{V} \), we can conclude that the concentration of the solution left in the container increases.

First, we need to determine the moles (n) of the salt present in the initial solution (0.50 M) with a mass of 4.5 grams. In order to do that, we can use the equation: \( C = \frac{n}{V} \) Since we want to find the moles of solute in the same volume of both solutions, we can rearrange the equation to solve for n: \[ n_{initial} = CV_{initial} \Longrightarrow n_{initial} = 0.50 M * V_{initial} \] Now we need to find out the mass-to-mole ratio of the salt in order to get the number of moles of salt in the initial solution.

We will use the mass of salt (4.5 g) in the 0.50 M solution to calculate the mass-to-mole ratio for the same volume: \[ mass\_to\_mole\_ratio = \frac{4.5 g}{n_{initial}} \]

We know the number of moles of solute in the same volume of a 2.50 M solution (n_final) can be calculated by: \[ n_{final} = CV_{final} \Longrightarrow n_{final} = 2.50 M * V_{final} \] Since V_initial = V_final, we can relate \(n_{initial}\) and \(n_{final}\) with the equation: \( \frac{n_{final}}{n_{initial}} = \frac{2.50 M}{0.50 M} \) Solving for \(n_{final}\) we get: \( n_{final} = 5 n_{initial} \) Using the mass-to-mole ratio, we can solve for the mass of salt in the 2.50 M solution (m_final): \[ m_{final} = mass\_to\_mole\_ratio * n_{final} = \frac{4.5 g}{n_{initial}} * 5 n_{initial} \] Cancelling out \(n_{initial}\) and simplifying, we get: \[ m_{final} = 22.5 g \] Thus, the mass of the salt present in the same volume of a 2.50 M solution is 22.5 grams.