Chapter 13: Problem 61
Calculate the molarity of each solution. (a) \(22.6 \mathrm{~g}\) of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in \(0.442 \mathrm{~L}\) of solution (b) \(42.6 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(1.58 \mathrm{~L}\) of solution (c) \(315 \mathrm{mg}\) of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in \(58.2 \mathrm{~mL}\) of solution
Short Answer
Step by step solution
Calculate the Molar Mass of C12H22O11
Calculate Moles of C12H22O11
Calculate Molarity of C12H22O11 Solution
Calculate the Molar Mass of NaCl
Calculate Moles of NaCl
Calculate Molarity of NaCl Solution
Convert Milligrams to Grams for C6H12O6
Calculate the Molar Mass of C6H12O6
Calculate Moles of C6H12O6
Convert Milliliters to Liters for C6H12O6
Calculate Molarity of C6H12O6 Solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass
For example, for sucrose (C12H22O11), the calculation involves multiplying the atomic mass of carbon (12.01 g/mol) by 12, the atomic mass of hydrogen (1.01 g/mol) by 22, and the atomic mass of oxygen (16.00 g/mol) by 11. Adding these together gives the molar mass of sucrose. Similarly, for sodium chloride (NaCl), the molar mass is found by adding the atomic weights of sodium (22.99 g/mol) and chlorine (35.45 g/mol).
Moles to Grams Conversion
The formula for converting grams to moles is simply: \[ \text{Number of moles} = \frac{\text{Mass in grams}}{\text{Molar mass}} \]. This relationship is used in stoichiometry calculations, where you often need to convert masses in grams to moles before performing further calculations.
Solution Concentration
Using the formula \[ Molarity (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \], students can determine the concentration of a solution. This is crucial for quantifying reactions in solution and for preparing solutions of desired concentrations in the laboratory.
Chemistry Stoichiometry
Stoichiometric calculations are essential for all kinds of chemistry applications, such as formulation of chemical reactions, laboratory experiments, and industrial processes. Understanding stoichiometry allows students to make detailed predictions about the outcomes of chemical reactions.
Molecular Formula
For instance, the molecular formula of glucose is C6H12O6, which informs us that each molecule is composed of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. The molecular formula is key in determining the molar mass, which in turn is used in molarity calculations and other stoichiometric computations.