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Chapter 9: Problem 47

Calculate the grams of solute needed to prepare each of the following: a. \(2.00 \mathrm{~L}\) of a \(1.50 \mathrm{M} \mathrm{NaOH}\) solution b. \(4.00 \mathrm{~L}\) of a \(0.200 \mathrm{M} \mathrm{KCl}\) solution c. \(25.0 \mathrm{~mL}\) of a \(6.00 \mathrm{M} \mathrm{HCl}\) solution

Short Answer

Expert verified
a. 120 g, b. 59.68 g, c. 5.475 g

Step by step solution

01

Formula for Molar Mass

Use the formula: \[ \text{{Molar mass}} (M) = \frac{{\text{{mass of solute (g)}}}}{{\text{{volume of solution (L)}} \times \text{{molarity (M)}}}} \]
02

Part a: Calculate the mass of NaOH

First, find the molar mass of NaOH. \[ \text{{Molar mass of NaOH}} = 23.0 + 16.0 + 1.0 = 40.0 \text{{ g/mol}} \] Given: Volume = 2.00 L Molarity = 1.50 M Calculate the mass: \[ \text{{mass (g)}} = \text{{Volume (L) }} \times \text{{Molarity (M) }} \times \text{{molar mass (g/mol) }} = 2.00 \text{{ L}} \times 1.50 \text{{ M}} \times 40.0 \text{{ g/mol}} = 120 \text{{ g}} \]
03

Part b: Calculate the mass of KCl

First, find the molar mass of KCl. \[ \text{{Molar mass of KCl}} = 39.1 + 35.5 = 74.6 \text{{ g/mol}} \] Given: Volume = 4.00 L Molarity = 0.200 M Calculate the mass: \[ \text{{mass (g)}} = \text{{Volume (L) }} \times \text{{Molarity (M) }} \times \text{{molar mass (g/mol) }} = 4.00 \text{{ L}} \times 0.200 \text{{ M}} \times 74.6 \text{{ g/mol}} = 59.68 \text{{ g}} \]
04

Part c: Calculate the mass of HCl

First, find the molar mass of HCl. \[ \text{{Molar mass of HCl}} = 1.0 + 35.5 = 36.5 \text{{ g/mol}} \] Given: Volume = 25.0 mL = 0.025 L Molarity = 6.00 M Calculate the mass: \[ \text{{mass (g)}} = \text{{Volume (L) }} \times \text{{Molarity (M) }} \times \text{{molar mass (g/mol) }} = 0.025 \text{{ L}} \times 6.00 \text{{ M}} \times 36.5 \text{{ g/mol}} = 5.475 \text{{ g}} \]

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Key Concepts

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

Solution Preparation
Preparing a solution involves dissolving a specific amount of solute in a solvent to achieve a desired concentration. This is essential in both laboratory and industrial settings. To accurately prepare a solution, you need to know two main things: the desired molarity (concentration) and the solution's volume.
Follow these steps for proper solution preparation:
  • Determine the mass of the solute required.
  • Measure the solvent volume accurately.
  • Dissolve the solute completely in the solvent.
  • Mix thoroughly to ensure uniform concentration.
Whether for small-scale experiments or large-scale production, ensuring the correct amounts and thorough mixing is key to reliable results.
Molarity
Molarity (M) is a way of expressing the concentration of a solution. It鈥檚 defined as the number of moles of a solute dissolved in one liter of solution. The formula to find molarity is:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Converting moles to grams often involves using the molar mass of the substance, which is the mass of one mole of that substance. For instance, to find the amount of solute needed for a certain molarity, you鈥檇 multiply the volume (in liters) by the molarity and the molar mass:
\[ \text{mass (g)} = \text{Volume (L)} \times \text{Molarity (M)} \times \text{Molar Mass (g/mol)} \]
This ensures the precise concentration suitable for experiments or reactions.
Grams of Solute
Calculating the grams of solute required for a solution involves using the molar mass, volume, and molarity. Here's a breakdown using our examples:
  • Part A: For NaOH, given a volume of 2.00 L and a molarity of 1.50 M, and knowing the molar mass of NaOH is 40.0 g/mol, we calculate the mass as:
    \[ \text{mass} = 2.00 \text{ L} \times 1.50 \text{ M} \times 40.0 \text{ g/mol} = 120 \text{ g} \]
  • Part B: For KCl, with a volume of 4.00 L and molarity of 0.200 M, and the molar mass of KCl is 74.6 g/mol, the mass is:
    \[ \text{mass} = 4.00 \text{ L} \times 0.200 \text{ M} \times 74.6 \text{ g/mol} = 59.68 \text{ g} \]
  • Part C: For HCl, with a volume of 25.0 mL (0.025 L) and a molarity of 6.00 M, plus the molar mass of HCl being 36.5 g/mol, the mass is:
    \[ \text{mass} = 0.025 \text{ L} \times 6.00 \text{ M} \times 36.5 \text{ g/mol} = 5.475 \text{ g} \]
By understanding these relationships, you ensure accurate solution preparation for precise experiments.

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Most popular questions from this chapter

How do temperature and pressure affect the solubility of solids and gases in water? (9.3)

Calculate the grams or milliliters of solute needed to prepare the following: a. \(50 . \mathrm{g}\) of a \(5.0 \%(\mathrm{~m} / \mathrm{m}) \mathrm{KCl}\) solution b. \(1250 \mathrm{~mL}\) of a \(4.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{NH}_{4} \mathrm{Cl}\) solution c. \(250 . \mathrm{mL}\) of a \(10.0 \%\) (v/v) acetic acid solution

Why would solutions with high salt content be used as food preservatives?

Calculate the final concentration of the solution when water is added to prepare each of the following: (9.5) a. \(25.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M}\) NaBr solution is diluted to \(50.0 \mathrm{~mL}\) b. \(15.0 \mathrm{~mL}\) of a \(12.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{K}_{2} \mathrm{SO}_{4}\) solution is diluted to \(40.0 \mathrm{~mL}\) c. \(75.0 \mathrm{~mL}\) of a \(6.00 \mathrm{M} \mathrm{NaOH}\) solution is diluted to \(255 \mathrm{~mL}\)

In a laboratory experiment, a 15.0 -mL sample of \(\mathrm{KCl}\) solution is poured into an evaporating dish with a mass of \(24.10 \mathrm{~g}\). The combined mass of the evaporating dish and \(\mathrm{KCl}\) solution is \(41.50 \mathrm{~g} .\) After heating, the evaporating dish and dry \(\mathrm{KCl}\) have a combined mass of \(28.28 \mathrm{~g}\). (9.4) a. What is the mass percent (m/m) of the KCl solution? b. What is the molarity (M) of the KCl solution? c. If water is added to \(10.0 \mathrm{~mL}\) of the initial \(\mathrm{KCl}\) solution to give a final volume of \(60.0 \mathrm{~mL},\) what is the molarity of the diluted KCl solution?
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