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Chapter 12: Problem 98

Define molarity.

Short Answer

Expert verified
Molarity is a measure of the concentration of a solute in a solution, expressed in moles of solute per liter of solution (mol/L), and denoted by the letter M. The formula for calculating molarity is \(M = \frac{n}{V}\), where \(M\) is the molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters. It is a crucial concept in chemistry for comparing concentrations of different solutions and performing calculations involving chemical reactions.

Step by step solution

01

Definition of Molarity

Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution (mol/L). It is often denoted by the letter M. This allows us to easily compare the concentrations of different solutions and perform calculations involving chemical reactions.
02

Formula for Calculating Molarity

The formula for calculating molarity (M) is given by the following equation: \[M = \frac{n}{V}\] where: - \(M\) is the molarity (mol/L) - \(n\) is the number of moles of solute (mol) - \(V\) is the volume of the solution (L)
03

Understanding Molarity

To better understand molarity, let's consider an example. Suppose we have 1 mole of sodium chloride (NaCl) dissolved in 1 liter of water. The molarity of this sodium chloride solution would be: \[M = \frac{1 \, \text{mol of NaCl}}{1 \, \text{L of solution}}\] Therefore, the molarity of this solution is 1 mol/L, or simply 1 M.
04

Using Molarity in Calculations

Molarity can be used in various calculations involving chemical reactions. For example, in a titration experiment, the molarity and volume of one reactant can be used to determine the molarity of another reactant: \[M_1V_1 = M_2V_2\] where: - \(M_1\) and \(M_2\) are the molarities of the two reactants - \(V_1\) and \(V_2\) are the volumes of the respective reactants In summary, molarity is a crucial concept in chemistry that helps us define the concentration of a solution and enables us to perform calculations involving chemical reactions.

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Most popular questions from this chapter

How many gallons of 24 -proof wine would you have to drink to consume \(0.100\) gallon of alcohol?

What is the volume in milliliters of \(0.0150 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to neutralize \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{MHNO}_{3}(a q) ?\)

A student dilutes \(45.0 \mathrm{~mL}\) of a \(0.500 \mathrm{M}\) solution of aluminum sulfate with sufficient water to prepare \(1.50 \mathrm{~L}\) of solution. (a) What is the molar concentration of aluminum sulfate in the diluted solution? Once in solution, the aluminum sulfate exists not intact but rather as dissociated ions. What are the molar concentrations (b) of \(\mathrm{Al}^{3+}(a q)\) in the diluted solution and (c) of \(\mathrm{SO}_{4}^{2-}(a q)\) in the diluted solution?

(a) How many grams of \(\mathrm{NaOH}\) are needed to prepare \(500.0 \mathrm{~mL}\) of a \(0.300 \mathrm{M} \mathrm{NaOH}\) solution? (b) Describe how you would make this solution, including the equipment needed.

A \(25.00 \mathrm{~mL}\) sample of aqueous sulfuric acid of unknown concentration is neutralized by \(27.55\) \(\mathrm{mL}\) of \(1.0002 \mathrm{M} \mathrm{NaOH}(a q)\) (a) Write a net ionic equation for this acid-base neutralization reaction. (b) How many moles of \(\mathrm{NaOH}\) did it take for the neutralization? (c) How many moles of sulfuric acid were neutralized? (d) What is the molar concentration of the sulfuric acid?
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