Question

If you dissolve 27 g of sugar into 155 \(\mathrm{mL}\) of water, what can you conclude about the mass and volume of the resulting solution? Assume a density of 1.00 \(\mathrm{g} / \mathrm{mL}\) for the water. a. \(\operatorname{mass}=155 \mathrm{g}\) ; volume \(=155 \mathrm{mL}\) b. \(\mathrm{mass}=182 \mathrm{g} ;\) volume \(=155 \mathrm{mL}\) c. mass \(=182\) g; volume \(>155 \mathrm{mL}\) d. \(\mathrm{mass}=155 \mathrm{g}\) ; volume \(>155 \mathrm{mL}\)

Short answer

The mass of the resulting solution is 182 g, and the volume remains at 155 mL, so the correct answer is option b.

Step-by-step solution

Determine the mass of the water

Given the density of water is 1.00 \(\mathrm{g} / \mathrm{mL}\), use the volume of the water to find its mass. Since the density is equal to mass divided by volume (D=m/V), and D equals 1.00 g/mL, the mass of water (m) in grams will be the same as its volume in milliliters. Therefore, mass of water = 155 g.

Calculate the total mass of the solution

To find the total mass of the solution, add the mass of the sugar to the mass of the water. The mass of sugar is 27 g. So the total mass of the solution is mass of water + mass of sugar, which equals 155 g + 27 g = 182 g.

Conclude the volume of the solution

Assuming the sugar dissolves completely and does not change the volume significantly, the volume of the water remains the same at 155 mL. In reality, the volume might increase slightly, but without data on the volume change due to sugar dissolution, we assume it's negligible.

Choose the correct option

Based on the calculated total mass and the assumption about the volume, the correct option is mass = 182 g and volume = 155 mL. Therefore, the correct answer is option b.

Key concepts

Mass-Volume Relationship

Understanding the mass-volume relationship is essential when working with solutions in chemistry. This relationship expresses how the mass of a substance relates to its volume, which is vitally important when mixing substances to create solutions. The key formula in these calculations is the density equation: \( Density = \frac{Mass}{Volume} \).
For water, with a known density of 1.00 \( \mathrm{g} / \mathrm{mL} \), this relationship simplifies. Given that the density is 1, the mass in grams of water equals its volume in milliliters.
When a substance like sugar is added to water, it contributes to the total mass without significantly changing the volume if the solute (sugar) completely dissolves. However, if you know the mass of the solute and the solvent (water), you can easily calculate the mass of the final solution by summing the two.

Solution Preparation

The preparation of solutions is a fundamental task in the lab. To accurately prepare a solution, one must know the desired concentration and the volume of the solvent to be used. Concentration is typically expressed in terms of mass per unit volume, such as \( \mathrm{g/mL} \) or \( \mathrm{mol/L} \).
In the case of dissolving sugar in water, assuming the sugar dissolves completely, the preparation is straightforward: measure the mass of the sugar, add it to the water, and ensure it mixes thoroughly.

Factors Influencing Solution Preparation

• The solute’s solubility: Determines how much can be dissolved.
• The temperature of the solvent: Can affect the amount of solute that dissolves.
• The presence of other substances: May influence the solubility and volume changes.

Ensuring that the sugar is fully dissolved without leaving undissolved particles is crucial for a homogenous solution and for the accuracy of the concentration and volume relationship.

Density of Water

Water has a standard density of 1.00 \( \mathrm{g/mL} \) at room temperature, which greatly simplifies calculations in chemistry. This unique property of water means that its mass in grams is numerically equivalent to its volume in milliliters. This characteristic of water is a baseline in many experimental setups and calculations.

Implications for Solution Calculations

When preparing solutions, the density of water allows you to easily determine the mass required for a specific volume. In some instances, adding solutes to water can change the volume slightly, but in many cases, such as with sugar in water at a low concentration, this change is negligible. This allows us to assume that the volume remains essentially the same after dissolution for practical purposes.