91Ó°ÊÓ

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air and others based on the high solubility of the heptahydrate \(\left(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right)\) form, also known as Epsom salt. The densities of the anhydrate and heptahydrate crystalline forms are 2.66 and \(1.68 \mathrm{g} / \mathrm{mL},\) respectively. Suppose you wish to form a 20.0 wt\% \(\mathrm{MgSO}_{4}\) aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at \(30^{\circ} \mathrm{C}\). The specific gravity of the 20.0 wt\% solution at \(30^{\circ} \mathrm{C}\) is \(1.22 .\) Answer the following questions for both forms of the \(\mathrm{MgSO}_{4}\) crystals: (a) What volume of water should be in the tank before crystals are added if the final product is to be 1000 kg of the 20 wt\% solution? (b) Suppose the tank diameter is \(0.30 \mathrm{m}\). What is the height of liquid in the tank before the crystals are added? (c) What is the height of the water in the tank after addition of the crystals but before they begin to dissolve? (d) What is the height of liquid in the tank after all the MgSO \(_{4}\) has dissolved?

Step by step solution

01

Calculate the Mass of MgSO4 in Solution

Firstly, we know that a 20 wt% solution means there are 20 grams of MgSO4 for every 100 g of solution. If, as the problem states, we want 1000 kg (or 1000000 g) of this solution, we need to find out how much MgSO4 this represents. We calculate this as follows: \(20\% \times 1000000 g = 200000 g = 200 kg \). This is the required MgSO4 (in either form) to be added.

02

Calculate the Amount of Water Required

The volume of water required before adding the crystals can be calculated using the weight of the MgSO4 and the weight of the solution. We already have the weight of the MgSO4 (200 kg), so we can subtract this from the total solution weight to get the amount of water necessary. We calculate this as follows: \(1000 kg (solution) - 200 kg (MgSO4) = 800 kg \). However, we're asked for the volume of water, not the weight. Given the density of water is approximately \(1 g/mL\), or \(1 kg/L\), this translates to \(800 L\) or \(0.8 m^3\).

03

Calculate the Height of the Water Before Addition

The height of water in the tank before the crystals are added corresponds to the volume of the water calculated in ‘Step 2’. Since the tank is cylindrical, the volume of the cylinder can be related to the height with the formula: \(V = πr²h\). Given the diameter (thus radius \(r = 0.3 m / 2 = 0.15 m\)) already, we can calculate the height \(h\) by rearranging the formula: \(h = V/(πr²) = 0.8m^3 / (π×(0.15m)²)\). Calculate this for the height in meters.

04

Calculate the Height of the Water After Addition

The height after addition of the MgSO4 crystals can be calculated by figuring out the volume of the crystals using their density, and then converting this to a height in the tank, in a similar process to Step 3. For each form, we'll divide the mass \(200 kg\) by the respective density (\(2.66 g/mL\) for the anhydrous form and \(1.68 g/mL\) for the heptahydrate). After converting these volumes from mL to the cubic meter, add each value to the initial volume of water (0.8 m^3), then use \(V = πr²h\) formula to determine the total height

05

Calculate the Height of the Liquid After Dissolution

The height of the liquid in the tank after all the MgSO4 has dissolved will return to the initial volume, because the mass hasn't been changed by dissolving the substances. Hence, the height remains the same as the end of the Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass and Volume Calculations

Understanding the relationship between mass, volume, and density is crucial when working with chemical solutions. The mass of a substance, measured in grams (g) or kilograms (kg), is its amount of matter. Volume, typically measured in liters (L) or milliliters (mL), is the space occupied by a substance. Density is a property that relates the mass of a substance to its volume, often expressed as g/mL or kg/L.

When we need to convert between mass and volume, we use the substance's density as a conversion factor. This is especially important in creating solutions where precise amounts of solutes and solvents are required. In the provided exercise, mass and volume calculations are employed to determine how much water and magnesium sulfate are needed to create a specific concentration of a solution.

Solution Preparation

Preparing a chemical solution involves dissolving a specific amount of solute (in this case, magnesium sulfate) in a solvent (water) to achieve a desired concentration. Such processes are guided by careful calculations and understanding of solubility.

In the given exercise, the aim is to prepare a 20 wt% aqueous solution of magnesium sulfate. It necessitates calculating the mass of the solute needed and the corresponding volume of the solvent. An in-depth understanding of solution preparation helps in accurately constructing chemical solutions vital for various applications, ensuring that the desired concentration is accurately achieved to maintain the consistency and integrity of an experiment or industrial process.

Aqueous Solutions

Aqueous solutions are formed when substances dissolve in water. Water's ability to dissolve a variety of substances makes it an excellent solvent for creating solutions in chemical processes. The solvency depends on the temperature and the physical properties of the solute, such as its solubility.

Creating a uniform aqueous solution, like the 20 wt% magnesium sulfate solution in the exercise, involves understanding the dissolution process. It's important to note that the properties of an aqueous solution can vary based on the concentration and the nature of the solute, and these solutions are commonly used in both laboratory settings and industrial applications.

Stoichiometry

Stoichiometry, at its core, deals with the quantification of reactants and products in a chemical reaction. It involves calculations that relate the quantities of substances involved in a reaction, grounded in the conservation of mass and the stoichiometric coefficients arising from the balanced chemical equations.

However, stoichiometry is not limited to reactions; it's also applied in the preparation of solutions, as seen with the magnesium sulfate exercise. Determining the right proportions of solute and solvent requires stoichiometric calculations. These calculations ensure that the desired concentration of a solution is achieved and are fundamental in a wide range of scientific disciplines, including chemistry, biology, environmental science, and engineering.