Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
molarity=molesofsolutelitersofsolution=mmolesofsolutemillilitersofsolution(4.5.1)
(4.5.1)molarity=molesofsolutelitersofsolution=mmolesofsolutemillilitersofsolution
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as MM . An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,
[sucrose]=1.00M(4.5.2)
(4.5.2)[sucrose]=1.00M
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
VLMmol/L=L(molL)=moles(4.5.3)
(4.5.3)VLMmol/L=L(molL)=moles
or
VmLMmmol/mL=mL(mmolmL)=mmoles(4.5.4)
(4.5.4)VmLMmmol/mL=mL(mmolmL)=mmoles
Figure 4.5.14.5.1 illustrates the use of Equations 4.5.34.5.3 and 4.5.44.5.4 .
alt
Figure 4.5.14.5.1 : Preparation of a Solution of Known Concentration Using a Solid Solute
Example 4.5.14.5.1 : Calculating Moles from Concentration of NaOH
Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.
Given: identity of solute and volume and molarity of solution
Asked for: amount of solute in moles
Strategy:
Use either Equation 4.5.34.5.3 or Equation 4.5.44.5.4 , depending on the units given in the problem.
Solution:
Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5.34.5.3 is more useful:
molesNaOH=VLMmol/L=(2.50L)(0.100molL)=0.250molNaOHmolesNaOH=VLMmol/L=(2.50L)(0.100molL)=0.250molNaOH
Exercise 4.5.14.5.1 : Calculating Moles from Concentration of Alanine
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
