The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed.
This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution.
In your case, you dissolve #"58.44 g"# of sodium chloride in #"1.0 L"# of water and make a #"1.0 M"# sodium chloride solution.
If you take
#100 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.1 L"#
of this solution, this sample must have the same concentration as the initial solution, i.e. #"1.0 M"#.
Notice that the volume of the sample is
#(1color(red)(cancel(color(black)("L"))))/(0.1color(red)(cancel(color(black)("L")))) = color(blue)(10)#
times smaller than the volume of the initial solution, so it follows that it must contain #color(blue)(10)# times fewer moles of solute in order to have the same molarity.
Sodium chloride was said to have a molar mass of #"58.44 g mol"^(-1)#, which tells you that #1# mole of sodium chloride has a mass of #"58.44 g"#.
Since the initial solution was made by dissolving the equivalent of
#"58.44 g " = " 1 mole NaCl"#
in #"10 L"# of water, it follows that the #"0.1 L"# sample must contain
#"1 mole NaCl"/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))#
This is equivalent to saying that the #"0.1 L"# sample contains
#"58.44 g NaCl"/color(blue)(10) = "5.844 g NaCl"#
ALTERNATIVELY
You can get the same result by using the formula for molarity, which is
#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#
Rearrange this equation to solve for #n_'solute"# and plug in your values to find
#c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"#
#n_"NaCl" = "1.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))#
Keep in mind that the volume must always be expressed in liters when working with molarity.
Definitions
Solute-the substance being dissolved
Solvent-the substance doing the dissolving (the larger amount)
Solution- a homogeneous mixture of the solute and the solvent
Solution= solvent + solute
Aqueous (aq)= water solution
Tincture= alcohol solution
Amalgam= Mercury solution
Molarity (M)- is the molar concentration of a solution measured in moles of solute per liter of solution.
The molarity definition is based on the volume of the solution, NOT the volume of water.
Vocab. Lesson
Incorrect= The solution is 5.0 Molarity.
Correct= The solution is 5.0 Molar.
Example Problems
Level 1- Given moles and liters
Determine the molarity when 3.0 moles of sucrose are dissolved to make 2.0 liters of solution.
| 3.0 mol | = X = 1.5M solution |
| 2.0 liters |
Level 2- Given Molarity and liters of solution
Determine the number of moles of salt dissolved in 5.0 liters of a 0.50M solution of salt water.
| X mol | = 0.5M solution |
| 5.0 liters |
cross multiply, X= 2.5 mols
Level 3- Given grams (instead of moles) and liters of solution
Determine the molarity when 117g of NaCl are dissolved to make 0.500 liters of solution.
1st convert to moles, 2nd plug into the molarity equation
117g NaCl( 1mol/58.5g)= 2.00mol NaCl
| 2.00 mol | = 4.00M solution |
| 0.500 liters |
Level 4-Given grams (instead of moles) and milliliters of solution (instead of liters)
Determine the molarity when 55.5g of CaCl2 are dissolved to make 250.mL of solution.
1st convert to moles, 2nd convert to liters, 3rd plug into the molarity equation
55.5g CaCl2( 1mol/111g)= 0.500mol CaCl2
250.ml( 1L/1000mL) =0.250L
| 0.500 mol | = 2.00M solution |
| 0.250 liters |
Past Regents Questions-Follow link to check the answers
Jan 2003-44 What is the molarity of a solution of NaOH if 2 liters of the solution contains 4 moles of NaOH?
(1) 0.5 M (3) 8 M
(2) 2 M (4) 80 M
Jan. 04-41 What is the molarity of a solution containing 20 grams of NaOH in 500 milliliters of solution?
(1) 1 M (2) 2 M (3) 0.04 M (4) 0.5 M
Jan 2002-42 What is the molarity of a solution that contains 0.50 mole of NaOH in 0.50 liter of solution?
(1) 1.0 M (3) 0.25 M
(2) 2.0 M (4) 0.50 M
Aug. 2006-42 How many total moles of KNO3 must be dissolved in water to make 1.5 liters of a 2.0 M solution?
(1) 0.50 mol (2) 2.0 mol (3) 3.0 mol (4) 1.3 mol
Aug 2005-
41 What is the total number of moles of NaCl(s) needed to make 3.0 liters of a 2.0 M NaCl solution?(1) 1.0 mol (3) 6.0 mol
(2) 0.70 mol (4) 8.0 mol
June 2006-
16 Molarity is defined as the(1) moles of solute per kilogram of solvent(2) moles of solute per liter of solution(3) mass of a solution
(4) volume of a solvent
Aug 2008-
15 Which phrase describes the molarity of a solution?(1) liters of solute per mole of solution(2) liters of solution per mole of solution(3) moles of solute per liter of solution
(4) moles of solution per liter of solution
June 2009-46 Which sample of HCl(aq) contains the greatest number of moles of solute particles?
(1) 1.0 L of 2.0 M HCl(aq)
(2) 2.0 L of 2.0 M HCl(aq)
(3) 3.0 L of 0.50 M HCl(aq)
(4) 4.0 L of 0.50 M HCl(aq)
June 2007-
13 A 3.0 M HCl(aq) solution contains a total of(1) 3.0 grams of HCl per liter of water(2) 3.0 grams of HCl per mole of solution(3) 3.0 moles of HCl per liter of solution
(4) 3.0 moles of HCl per mole of water
June 2010-14 The molarity of an aqueous solution of NaCl is defined as the(1) grams of NaCl per liter of water(2) grams of NaCl per liter of solution(3) moles of NaCl per liter of water
(4) moles of NaCl per liter of solution
Jan 2008-
15 Which unit can be used to express solution concentration?(1) J/mol (3) mol/L
(2) L/mol (4) mol/s
Jan 04-41 What is the Molarity of a solution containing 20 grams of NaOH in 500 milliliters of solution?(1) 1 M (3) 0.04 M
(2) 2 M (4) 0.5 M
Jan 2010-40 What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass =26 grams/mole)?(1) 1.3 M (3) 3.0 M
(2) 2.0 M (4) 0.75 M
on to ppm or Molality
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