# Chemical Reactions Ch

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```					                            Ch.11 (Continued)

Water and Solutions

 Solutions
 Solution Concentrations
o % By Weight
o Molarity
I. Solutions
A solution is a mixture of two
substances.

Solvent                      Solute
Largest in amount       smallest in amount
Examples: Sugar solution: solute = sugar
Solvent = water

II. Solution Concentration:
Relative amounts of solute and solvent are described by the
concentration, of a solution. A solution with a large amount of
solute is called concentrated. A solution with much less solute is
called a dilute solution.

Two types of concentrations, we are primarily interested in this
course are:

% by weight                                                   Molarity
(M)
% by weight:
This is defined as the mass fraction multiplied by 100%.

mass of solute
% by weight                            100%
mass of solution
Q. Vinegar is a mixture of acetic acid in water, usually 5% by
weight. How many grams of pure acetic acid are in 25.0 g of
vinegar?

Given:    % by weight of solute = 5%
Mass of solution = 25.0 g
Mass of solute = ?

From the above equation we get

mass of        solution % by weight
mass of        solute 
100%     solution

25 g    solution 5%     solute

100%     solution
25  5
        g solute
100
 1.25 g   solute

Q. Sterile saline solutions containing NaCl in water are often
used in medicine. What is the percent by weight of NaCl in a
solution made by dissolving 4.6 g NaCl in 500 g pure water?

Q. What is the weight percent of glucose in a solution containing
21.5 g glucose in 750 g pure water?
Molarity
The molarity of a solution is defined as

number of moles of                  solute
Molarity 
number of liters of                 solution

From this we also get that the number of moles of solute, which is

moles of        solute  Molarity  volume of              solution      (L)

Q. Calcuate the number of moles of KNO3 in 250.0mL of 0.020M
KNO3 solution.

0.020 mole      KNO
0.250 L                           3
 5.0  10 mole
3
KNO      3
1L
Q. A student dissolves 36g Na2SO4 in water to make a 750.0 mL
solution. Calculate the molarity of the sodium sulfate in the
solution.

1mole Na SO
36 g      Na SO 
2   4
2    4

142 g  Na SO
molarity                                           2   4

1L
750 mL 
10 mL       3

36 10 mole 3


142  750 L
 0.338 M

Q. Calculate the mass in grams of NiCl2 needed to prepare 500.0
mL of 0.125 M NiCl2.

Q. Calculate the mass in grams of NaBr needed to prepare 250.0
mL of 0.075 M NaBr.

Q. Calculate the molarity of a soute in a solution containing
(a) 14.2 g KCl in 250.0 mL solution
(b) 2.20 g (NH4)2SO4 in 400.0 mL solution
(c) 15.0 g C6H12O6 in 500.0 mL solution

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