CHAPTER FOUR
Molarity
4.78
The equation for calculating molarity is:
M = molarity = moles of solute
liters of soln
The equation to calculate molarity is moles of solute per liter of solution. It is a convenient quantity to
use because it involves to easily measured quantities. One can weigh out a known mass of a compound
and prepare a solution of a known volume with ease.
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The steps involved in preparing a solution of known molar concentration using a volumetric flask are as
follows:
1. The solute is accurately weighed and transferred to a volumetric flask through a funnel.
2. Water is then added to the flask, which is carefully swirled to dissolve the solid.
3. After all the solid has dissolved, more water is added slowly to bring the level of solution
exactly to the volume mark.
4. Knowing the volume of the solution in the flask and the quantity of compound dissolved,
we can calculate the molarity of the solution.
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First, calculate the moles of KI needed to prepare the solution. (5.00 × 102 mL = 0.500 L)
mol = M × L
mol KI =
2.80 mol KI
× 0.500 L soln = 1.40 mol KI
1 L soln
Converting to grams of KI:
1.40 mol KI ×
4.81
166.0 g KI
= 232 g KI
1 mol KI
Step 1: From the molarity (0.707 M), we can calculate the moles of NaNO3 needed to prepare 250 mL
of solution.
0.707 mol NaNO3
1 L soln
Moles NaNO3 =
×
× 250 mL soln = 0.177 mol
1 L soln
1000 mL soln
Step 2: Use the molar mass of NaNO3 as a conversion factor to convert from moles to grams.
M (NaNO3) = 85.00 g/mol.
0.177 mol NaNO3 ×
4.82
85.00 g NaNO3
= 15.0 g NaNO 3
1 mol NaNO3
mol = M × L
60.0 mL = 0.0600 L
mol MgCl2 =
4.83
0.100 mol MgCl2
× 0.0600 L soln = 6.00 × 10−3 mol MgCl2
1 L soln
Since the problem asks for grams of solute (KOH), you should be thinking that you can calculate moles
of solute from the molarity and volume of solution. Then, you can convert moles of solute to grams of
solute.
? moles KOH solute =
5.50 moles solute
× 0.0350 L solution = 0.193 moles KOH
1 L solution
The molar mass of KOH is 56.11 g/mol. Use this conversion factor to calculate grams of KOH.
? grams KOH = 0193
.
mol KOH ×
4.84
Molar mass of C2H5OH = 46.07 g/mol; molar mass of C12H22O11 = 342.3 g/mol; molar mass of
NaCl = 58.44 g/mol.
(a)
(b)
(b)
mol solute
0.154 mol NaCl
= 1.78 M
=
L of soln
86.4 × 10−3 L soln
1 mol CaCl 2
= 0.0937 mol CaCl2
111.0 g CaCl 2
0.0937 mol CaCl 2
= 0.426 M
0.220 L
? mol C10H8 = 7.82 g C10 H 8 ×
M =
1 mol CH 3OH
= 0.205 mol CH3OH
32.04 g CH 3OH
0.205 mol CH 3OH
= 1.37 M
0150
.
L
? mol CaCl2 = 10.4 g CaCl 2 ×
M =
(c)
1 mol NaCl
= 0.154 mol NaCl
58.44 g NaCl
? mol CH3OH = 6.57 g CH3OH ×
M =
1 mol C12 H 22 O11
= 0.0450 mol C12H22O11
342.3 g C12 H 22 O11
0.0450 mol C12 H 22 O11
mol solute
= 0.608 M
=
L of soln
74.0 × 10−3 L soln
? mol NaCl = 9.00 g NaCl ×
Molarity =
(a)
0.629 mol C2 H5 OH
mol solute
=
= 1.15 M
L of soln
0.545 L soln
? mol C12H22O11 = 15.4 g C12H22O11 ×
Molarity =
(c)
1 mol C2 H5 OH
= 0.629 mol C2H5OH
46.07 g C2 H5 OH
? mol C2H5OH = 29.0 g C2H5OH ×
Molarity =
4.85
56.11 g KOH
= 10.8 g KOH
1 mol KOH
1 mol C10 H 8
= 0.0610 mol C10H8
128.2 g C10 H 8
0.0610 mol C10 H 8
= 0.716 M
0.0852 L
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First, calculate the moles of each solute. Then, you can calculate the volume (in L) from the molarity
and the number of moles of solute.
(a)
? mol NaCl = 2.14 g NaCl ×
L soln =
(b)
1 mol C2 H5 OH
= 0.0933 mol C2H5OH
46.07 g C2 H5 OH
0.0933 mol C2 H5 OH
mol solute
=
= 0.0622 L = 62.2 mL soln
Molarity
1.50 mol / L
? mol CH3COOH = 0.85 g CH3COOH ×
L soln =
4.87
mol solute
0.0366 mol NaCl
=
= 0.136 L = 136 mL soln
Molarity
0.270 mol / L
? mol C2H5OH = 4.30 g C2H5OH ×
L soln =
(c)
1 mol NaCl
= 0.0366 mol NaCl
58.44 g NaCl
1 mol CH3COOH
= 0.014 mol CH3COOH
60.05 g CH3COOH
0.014 mol CH3COOH
mol solute
=
= 0.047 L = 47 mL soln
Molarity
0.30 mol / L
A 250 mL sample of 0.100 M solution contains 0.0250 mol of solute (mol = M × L). The computation in
each case is the same:
259.8 g CsI
= 6.50 g CsI
1 mol CsI
(a)
0.0250 mol CsI ×
(b)
0.0250 mol H 2SO 4 ×
(c)
0.0250 mol Na 2 CO 3 ×
(d)
0.0250 mol K 2 Cr2 O 7 ×
(e)
0.0250 mol KMnO 4 ×
98.09 g H 2SO 4
= 2.45 g H2SO4
1 mol H 2 SO 4
106.0 g Na 2 CO 3
= 2.65 g Na2CO3
1 mol Na 2 CO 3
294.2 g K 2 Cr2 O 7
= 7.36 g K2Cr2O7
1 mol K 2 Cr2 O 7
158.0 g KMnO 4
= 3.95 g KMnO4
1 mol KMnO 4