Chapter 14: Problem 60
What is the mass of solute in each of the following solutions? (a) 1.00 L of \(0.100 \mathrm{M} \mathrm{NaOH}\) (b) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{LiHCO}_{3}\) (c) \(25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{CuCl}_{2}\) (d) \(25.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KMnO}_{4}\)
Short Answer
Expert verified
(a) 4.00 g NaOH
(b) 6.10 g LiHCO3
(c) 1.68 g CuCl2
(d) 1.98 g KMnO4
Step by step solution
01
Calculate Moles of Solute for NaOH
First, determine the moles of NaOH in 1.00 L of a 0.100 M solution. Use the formula: \( \text{moles} = \text{Molarity} \times \text{Volume (L)} \). Here, \( \text{moles} = 0.100 \times 1.00 = 0.100 \text{ moles of NaOH} \).
02
Convert Moles to Mass for NaOH
Find the molar mass of NaOH, which is 23 (Na) + 16 (O) + 1 (H) = 40 g/mol. Use the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \). Thus, \( \text{mass} = 0.100 \times 40 = 4.00 \text{ g of NaOH} \).
03
Calculate Moles of Solute for LiHCO3
Determine the moles of LiHCO3 in 1.00 L of a 0.100 M solution: \( \text{moles} = 0.100 \times 1.00 = 0.100 \text{ moles of LiHCO}_{3} \).
04
Convert Moles to Mass for LiHCO3
Find the molar mass of LiHCO3, which is 7 (Li) + 1 (H) + 12 (C) + 3×16 (O) = 61 g/mol. Then calculate the mass: \( \text{mass} = 0.100 \times 61 = 6.10 \text{ g of LiHCO}_{3} \).
05
Calculate Moles of Solute for CuCl2
Convert 25.0 mL to liters: \( \text{Volume} = 0.025 \text{ L} \). Then, find moles using: \( \text{moles} = 0.500 \times 0.025 = 0.0125 \text{ moles of CuCl}_{2} \).
06
Convert Moles to Mass for CuCl2
The molar mass of CuCl2 is 63.5 (Cu) + 2×35.5 (Cl) = 134.5 g/mol. Calculate the mass: \( \text{mass} = 0.0125 \times 134.5 = 1.68 \text{ g of CuCl}_{2} \).
07
Calculate Moles of Solute for KMnO4
Convert 25.0 mL to liters: \( \text{Volume} = 0.025 \text{ L} \). Then, find moles using: \( \text{moles} = 0.500 \times 0.025 = 0.0125 \text{ moles of KMnO}_{4} \).
08
Convert Moles to Mass for KMnO4
The molar mass of KMnO4 is 39 (K) + 55 (Mn) + 4×16 (O) = 158 g/mol. Calculate the mass: \( \text{mass} = 0.0125 \times 158 = 1.98 \text{ g of KMnO}_{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass
Molar mass is a fundamental concept that allows us to connect the mass of a substance to the amount of substance present, measured in moles. It is defined as the mass of one mole of a given substance, usually expressed in grams per mole (g/mol).
To calculate the molar mass, you must add up the atomic masses of all the atoms in a molecule. These atomic masses are usually found on the periodic table. For example, the molar mass of sodium hydroxide (NaOH) is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H): 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
Understanding molar mass helps in converting between grams and moles in chemical equations and reactions, playing a crucial role in stoichiometry and chemical solutions. Always remember to use the correct number of significant figures when performing calculations, as this ensures precision and accuracy.
To calculate the molar mass, you must add up the atomic masses of all the atoms in a molecule. These atomic masses are usually found on the periodic table. For example, the molar mass of sodium hydroxide (NaOH) is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H): 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.
Understanding molar mass helps in converting between grams and moles in chemical equations and reactions, playing a crucial role in stoichiometry and chemical solutions. Always remember to use the correct number of significant figures when performing calculations, as this ensures precision and accuracy.
Chemical Solutions
Chemical solutions are mixtures where a solute is dissolved into a solvent, creating a homogenous mixture. The most common solvent is water, making aqueous solutions.
When working with solutions, one often refers to the concentration of the solute, commonly measured in units like molarity (M), which is moles of solute per liter of solution.
A detailed understanding of chemical solutions involves knowing how to prepare them and calculate their concentrations. This often involves dissolving a known mass of solute and then diluting it to the desired volume. Calculating the mass of solute needed for a particular molarity can be done using formulae such as: \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{Liters of solution}}\]Let's say you need to make a 0.500 M solution of copper(II) chloride (CuCl₂), and you only have 25.0 mL of solution. You first convert the volume to liters (0.025 L) and use the molarity formula to find the moles of CuCl₂, which can then be converted to grams using its molar mass.
When working with solutions, one often refers to the concentration of the solute, commonly measured in units like molarity (M), which is moles of solute per liter of solution.
A detailed understanding of chemical solutions involves knowing how to prepare them and calculate their concentrations. This often involves dissolving a known mass of solute and then diluting it to the desired volume. Calculating the mass of solute needed for a particular molarity can be done using formulae such as: \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{Liters of solution}}\]Let's say you need to make a 0.500 M solution of copper(II) chloride (CuCl₂), and you only have 25.0 mL of solution. You first convert the volume to liters (0.025 L) and use the molarity formula to find the moles of CuCl₂, which can then be converted to grams using its molar mass.
Stoichiometry
Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It is based on the mole concept and balanced chemical equations.
When you balance a chemical equation, it provides the necessary ratios of reactants to products. This ratio allows you to convert between mass, moles, and molecules. For example, stoichiometry enables you to calculate how much of a reactant is required to completely react with another reactant or how much product will be generated.
Using stoichiometry involves setting up conversions between units using conversion factors: moles-to-moles (using coefficients from balanced equations), moles-to-mass (using molar mass), and sometimes even moles-to-volume (using 22.4 L/mol at STP for gases). Always start with a balanced chemical equation. From there, you'll follow a series of conversion steps, adjusting for the quantities required or produced in your reaction. This understanding is essential, especially in industrial chemical applications, where precise calculations affect the economy and safety of processes.
When you balance a chemical equation, it provides the necessary ratios of reactants to products. This ratio allows you to convert between mass, moles, and molecules. For example, stoichiometry enables you to calculate how much of a reactant is required to completely react with another reactant or how much product will be generated.
Using stoichiometry involves setting up conversions between units using conversion factors: moles-to-moles (using coefficients from balanced equations), moles-to-mass (using molar mass), and sometimes even moles-to-volume (using 22.4 L/mol at STP for gases). Always start with a balanced chemical equation. From there, you'll follow a series of conversion steps, adjusting for the quantities required or produced in your reaction. This understanding is essential, especially in industrial chemical applications, where precise calculations affect the economy and safety of processes.
