Chemistry Reference
In-Depth Information
moles of solute
kilograms of solvent
4.
[molality]—
Molality =
5.
[molarity]—We read M as “molar.”
6.
[molality]—We read m as “molal.”
moles of solute
liters of solution
2.0 moles
3.0 L
7.
[0.67 M]—
Molarity (M) =
=
= 0.67 M
8.
[100. g]—First, find out how many moles of NaOH you need. Then multiply by
the molar mass of NaOH, which is (23.0 g + 16.0 g + 1.01 g) = 40.0 g/mole.
# moles of solute = molarity × liters of solution = 1.25 M × 2.00 L = 2.50 moles
Mass = # of moles × molar mass = 2.50 moles × 40.0 g/mole = 100. g
liters of solution =
moles of solute
molarity
8.0 moles
2.0 M
9.
[4.0 L]—
=
= 4.0 L
10. [0.5 kg]—The molar mass of CaCl
2
is 111.1 g (40.1 g + 71.0 g = 111.1 g), so
you start with 1.00 mole of it. The kilograms of solvent that you need can be
determined with the following formula:
moles of solute
molality
1.0 moles
2.0 m
kilograms of solvent =
=
= 0.5 kg
Lesson 9-2 Review
1. [acid]—A common method for producing hydrogen gas in a chemistry lab
involves zinc reacting with an acid, such as HCl, in a single replacement
reaction.
2. [acid]—Litmus paper is a common indicator found in the chemistry lab.
3. [base]—Neutral substances have pH values of 7; above that is considered
basic.
4. [acid]—Of course, you would never taste a strong or an unknown acid. We
are talking about weak acids such as lemon juice.
5. [base]—Sodium hydroxide is found in some types of drain cleaners.
6. [base]—Remember: Litmus paper is only one of several types of indicators
that you may encounter as you study chemistry.
7. [5]—We solve for pOH as shown here:
pOH = -log[OH
-
] = -log(1 × 10
-5
) = 5
8. [2.2]—This time, you really need to use the formula. Practice using your
calculator correctly, by making sure that you can find this answer.
pH = -log[H
3
O
+
] = -log(5.9 × 10
-3
) = 2.229147988 = 2.2
9. [9.5]—Again, you are given the concentration of the hydroxide ions, rather
than the concentration of the hydronium ions. The easiest way to solve this
is shown.
pH = 14 - pOH = 14 - (-log[OH-]) = 14 - (-log(3.5 × 10
-5
)) = 9.544068044 = 9.5
10. [1 × 10
-5
M]—See the Ion Concentrations chart on page 306.
