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Calculate the amount of water (in grams) that must be added to (a) \(5.00 \mathrm{~g}\) of urea \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) in the preparation of a 16.2 percent by mass solution, and (b) \(26.2 \mathrm{~g}\) of \(\mathrm{MgCl}_{2}\) in the preparation of a 1.5 percent by mass solution.

Short Answer

Expert verified
(a) 25.86 g of water; (b) 1720.47 g of water.

Step by step solution

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01

Understand the formula for percent by mass

The percent by mass (or mass percent) formula is: \[ \text{Percent by Mass} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 \] We need to rearrange this formula to calculate the total mass of the solution and subsequently the mass of water needed.
02

Solve for 5.00 g of Urea

Given: \( 5.00 \, \text{g} \) of \( \text{urea} \) is to be in a 16.2% by mass solution.Using the formula for percent by mass: \[ 16.2 = \left( \frac{5.00}{\text{Mass of solution}} \right) \times 100 \]Rearrange to find the mass of the solution:\[ \text{Mass of solution} = \frac{5.00 \times 100}{16.2} \approx 30.86 \, \text{g} \]Calculate the amount of water needed:\[ \text{Mass of water} = \text{Mass of solution} - \text{Mass of urea} = 30.86 - 5.00 = 25.86 \, \text{g} \]
03

Solve for 26.2 g of MgCl2

Given: \( 26.2 \, \text{g} \) of \( \text{MgCl}_{2} \) in a 1.5% by mass solution.Using the formula for percent by mass: \[ 1.5 = \left( \frac{26.2}{\text{Mass of solution}} \right) \times 100 \]Rearrange to find the mass of the solution:\[ \text{Mass of solution} = \frac{26.2 \times 100}{1.5} \approx 1746.67 \, \text{g} \]Calculate the amount of water needed:\[ \text{Mass of water} = \text{Mass of solution} - \text{Mass of MgCl}_{2} = 1746.67 - 26.2 = 1720.47 \, \text{g} \]
04

Conclusion

Based on our calculations:- (a) For the urea solution, we need to add approximately \( 25.86 \, \text{g} \) of water.- (b) For the \( \text{MgCl}_{2} \) solution, we need to add approximately \( 1720.47 \, \text{g} \) of water.

Key Concepts

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

Percent by Mass
The concept of "percent by mass" is a way to express the concentration of a particular substance in a mixture or solution. It specifically denotes the mass of the solute divided by the total mass of the solution, multiplied by 100 to convert it into a percentage. This value helps in understanding how much solute is present in a given quantity of solution.

The formula for percent by mass is given as:
  • \[ \text{Percent by Mass} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 \]
This method is commonly used in chemistry to create solutions with precise concentrations.

It is essential for various applications such as preparing medication doses or chemical reactions to ensure optimal conditions.
Mass of Solute
The mass of the solute is simply the weight of the substance that is dissolved in the solution. In the context of making solutions, the solute is the component present in a smaller amount compared to the solvent. For the given exercises, urea and MgCl2 are the solutes.

In a practical scenario, the mass of the solute needs to be known before starting to prepare a solution of a specific concentration. Knowing the mass of the solute helps in applying the percent by mass formula to find out other values, like the mass of the entire solution or the amount of solvent needed.

In exercises, it's crucial to convert all masses to the same unit, typically grams, to maintain consistency and accuracy in calculations.
Mass of Solution
The mass of the solution is the total weight of both the solute and the solvent combined. This value gives a complete picture of the solution's entirety.

In calculations, the mass of the solution is typically used to determine the required mass of solvent needed to achieve a desired solution concentration. This is especially important in scientific experiments where precise concentrations are needed to ensure accurate and reproducible results.

To find the mass of the solution, rearrange the percent by mass formula to solve for it. This step is crucial in chemistry exercises, as it bridges the information between the amount of solute and the concentration of the solution.
Urea
Urea, chemically known as \( \text{NH}_2\text{CONH}_2 \), is a common organic compound used in fertilizers and various industrial applications. When considering it in a solution, urea acts as the solute that imparts certain properties to the mixture.

When preparing a solution with urea, determining how much water or another solvent to add is essential to reach a specific concentration. For instance, creating a solution that is 16.2% urea by mass requires careful measurement of both urea and water.

It's important to handle urea solutions carefully, as the concentration affects its reactivity and efficacy in fertilizers.
MgCl2
Magnesium chloride (MgCl2) is an ionic compound widely used in various industries, including the production of magnesium metal, de-icing roads, and in the medical field. In a solution form, MgCl2 typically serves as the solute, contributing to the mixture's properties.

To make a solution with a specific percent by mass of MgCl2, accurate measurement of MgCl2 and water is crucial. For example, to prepare a 1.5% by mass solution, you need to calculate the mass of water to mix with a known mass of MgCl2.

Understanding the concentration of such solutions is important for their respective applications, ensuring they are effective and safe for use.

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Most popular questions from this chapter

Which of the following aqueous solutions has (a) the higher boiling point, (b) the higher freezing point, and (c) the lower vapor pressure: \(0.35 \mathrm{~m} \mathrm{CaCl}_{2}\) or \(0.90 \mathrm{~m}\) urea? Explain. Assume complete dissociation.

The solubility of \(\mathrm{N}_{2}\) in blood at \(37^{\circ} \mathrm{C}\) and at a partial pressure of 0.80 atm is \(5.6 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\). A deep-sea diver breathes compressed air with the partial pressure of \(\mathrm{N}_{2}\) equal to \(4.0 \mathrm{~atm}\). Assume that the total volume of blood in the body is \(5.0 \mathrm{~L}\). Calculate the amount of \(\mathrm{N}_{2}\) gas released (in liters at \(37^{\circ} \mathrm{C}\) and \(\left.1 \mathrm{~atm}\right)\) when the diver returns to the surface of the water, where the partial pressure of \(\mathrm{N}_{2}\) is \(0.80 \mathrm{~atm}\).

Determine the van't Hoff factor of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) in a \(0.40-\mathrm{m}\) solution whose freezing point is \(-2.6^{\circ} \mathrm{C}\).

(a) The root cells of plants contain a solution that is hypertonic in relation to water in the soil. Thus, water can move into the roots by osmosis. Explain why salts such as \(\mathrm{NaCl}\) and \(\mathrm{CaCl}_{2}\) spread on roads to melt ice can be harmful to nearby trees. (b) Just before urine leaves the human body, the collecting ducts in the kidney (which contain the urine) pass through a fluid whose salt concentration is considerably greater than is found in the blood and tissues. Explain how this action helps conserve water in the body.

A student is observing two beakers of water. One beaker is heated to \(30^{\circ} \mathrm{C},\) and the other is heated to \(100^{\circ} \mathrm{C}\). In each case, bubbles form in the water. Are these bubbles of the same origin? Explain.

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