Ch 13 Solutions Sec 2 pg 467 #1-11

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Ch 13 Solutions Sec 2
pg 467 #1-11
1. Why did chemists develop
the concept of molarity?
• To easily connect the amount in moles
of a substance with its concentration in
a solution
2. In what units is molarity
expressed
• Moles per liter
• Mol/L
3. Describe in your own
words how to prepare 100.0
mL of a 0.85 M solution of
sodium chloride. 23Na + 36Cl= 59NaCl
• .85 mol NaCl x 59 g/mol = 50 g/L or
5.0g NaCl for 1/10 L (100ml)
• Weigh 5.0 g of salt, dissolve in water.
Pour into 100 ml volumetric flask, swirl
and fill to volumetric line
4. If you dissolve 2.00 mol KI
in 1.00 L of water, will you get
2.00 M solution? Explain.
• No
• The total volume may be different by the
time the solute is dissolved.
5. A sample of 400.0 g of water
is found to contain 175 mg Cd.
What is this concentration in
parts per million.
•
•
•
•
Change mg to g
175 mg x 1 g/ 1000 mg = .175 g Cd
Find g Cd/g of water
.175 g/ 400.0 g water x 1000,000 parts
/1 million =
• 437.5 or 438 parts per million or ppm
6. If 1.63 x 10-4 g of helium
dissolves in 100.0 g of water, what
is the concentration in parts per
million?
• How much in 1 gram of water?
• 1.63 x 10-4 g He / 100g H2O x 1000,000 parts/million
• 1.63 parts per million or ppm He
A standard solution of NaOH is
1.000 M. What mass of NaOH
is present in 100.0 mL of the
solution?
•
•
•
•
Na = 23, O = 16, H = 1 NaOH = 40
40 g in 1 M
100 mL is 1/10 L
so 40/10 = 4 g in 100.0 mL
8. A 32 g sample of LiCl is
dissolved in water to to form
655 mL of solution. What is the
molarity of the solution?
• Convert mass to moles. Li = 6.9,Cl = 36
• 32 g x mol/43g LiCl = .74 mol LiCl
• Convert mL to L
• 655 ml x 1 L/ 1000mL = .655 L
• Find Moles per Liter
.74 mol / .655 L = 1.129 or 1.1 M LiCl
9. Most household bleach
contains sodium hypochlorite,
NaOCl. A 2.84 L bottle
contains 177 g NaOCl. What is
the molarity of the solution.
• Find moles of NaOCl Na=23,O=16,Cl=35.5
• 177g NaOCl x mol/74.5g NaOCl = 2.38 Mol
NaOCl
• Find mol/L
• 2.38 mol/2.84 L = .8365 or .837 mol/L or M
10. What mass of AgNO3 is
needed to prepare 250.0 mL
of a 0.125 M solution.
Ag=108,N=14,O3= 16(3)= 48 AgNO3 = 170 g
M = g/L = 170g /mol = for 1 Mol solution
170/4 for .250 L = 42.5 g for .250 L AgNO3
42.5 g for 1 M, 42.5 x .125 M
5.25 or 5.3 g AgNO3 for .125 M solution
in the reaction described by the following equation:
2H3PO4(aq) + 3Ca(OH)2(aq) -->
Ca3(PO4)2(s) + 6H2O(aq)
Ca=40(3), P=31(2), O=16(8)=128, = 310g total
H= 1(2), O=16, total 18g total
What mass in grams of each product would be formed
if 7.5 L of 5.00 M phosphoric acid reacted with an
excess of calcium hydroxide?
• H3PO4 = 1(3) + 31 + 16(4) or 64 = 98g for 1 M
• 98x5 x 7.5 = 3,562.5 or 3600 g H3PO4
• 3600g 2H3PO4 x 1 mol/98g x 1 mol Ca3(PO4)2 /2mol 2H3PO4
• x310 g Ca3(PO4)2 /1 mol Ca3(PO4)2 = 5,693.8 or 5.7 x 103 Ca3(PO4)2
• 18g H2O
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