Molarity Problems Part Deux

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• Molarity Calculations
Calculate the molarities of the following solutions:

1) 2.3 moles of sodium chloride in 0.45 liters of water.

(2.3 moles) / (.45 liters)  =  5.1 moles/liter

2) 1.2 moles of calcium carbonate in 1.22 liters of water.

(1.2 moles) / (1.22 liters) = .984 moles/liter

3) 0.09 moles of sodium sulfate in 12 mL of water.

(0.09 moles) / (.012 liters) = 7.5 moles/liter

4) 0.75 moles of lithium fluoride in 65 mL of water.

(.75 moles) / (.064 liters) = 11.7 moles/liter

5) 0.8 moles of magnesium acetate in 5 liters of water.

(.8 moles) / (5 liters) = .16 moles/liter

6) 120 grams of calcium nitrite in 240 mL of water.

• Atomic mass of Ca(NO2)2 = 20.08 g/mole + (14.01 g/mole)(2) + (16.00 g/mole)(4) = 112.1 g/mole
• (120 g)/(112.1 g/mole) = 1.07 moles
• (1.07 moles) / (.240 liters) = 4.46 moles/liter

7) 98 grams of sodium hydroxide in 2.2 liters of water.
8) 1.2 grams of hydrochloric acid in 25 mL of water.
9) 45 grams of ammonia in 0.75 L of water.

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Explain how you would make the following solutions. You should tell how many
grams of the substance you need to make the solution, not how many moles.

10) 2 L of 6 M HCl

• 1 liter of a 6 M solution means there are 6 moles of HCl per liter, so...
• the atomic mass of HCl is (1.008 g/mole) + (35.45 g/mole) = 36.45 g/mole
• 1 mole of HCl is:  (1 mole)(36.45 g/mole) = 36.45 grams
• However, we don't want 1 mole, we want 6 moles, so we increase our recipe by 6:  6 x 36.45 grams = 218.7 g
• We're still not done though, we need twice as much solution (2 liters) so we increase our recipe by 2:  (218.7g)(2) = 437.4g

6.0 moles HCl 36.45 g HCl 2 liters
1 liter
1 mole HCl

= 437.4 grams HCl (ack sig figs are messed up here...)

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11) 1.5 L of 2 M NaOH

• Let's keep it simple and make 1 liter first so we need 2 moles / liter
• NaOH = (22.99 g/mole) + (16.00 g/mole) + (1.008 g/mole) = 40.00 g/mole
• We want 2 moles so (40.00 g/mole)(2 mole) = 80.00 grams
• So to make 1 liter of 2M NaOH takes 80.00 grams so to make 1.5 liter we would need 1.5 times as much NaOH ==>  80.00 grams x 1.5 = 120.0 grams

2 moles NaOH 40.00 g NaOH 1.5 liters
1 liter
1 mole NaOH

=120 grams NaOH

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12) 0.75 L of 0.25 M Na2SO4

• We want a .25 M solution, so let's make 1 liter first
• We need the atomic mass first:  (22.99 g/mole)(2) + (32.07 g/mole) + (16.00 g/mole)(4) = 142.1 g/mole
• (.25 moles)(142.1 grams/mole) = 35.51 grams
• However, that is for 1 liter of solution and we want .75 liters so...
• (.75)(35.51 grams) = 26.63 grams
.25 moles Na2SO4 142.1 g Na2SO4 .75 liters
1 liter
1 mole Na2SO4

= 26.64 grams of Na2SO4

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13) 45 mL of 0.12 M sodium carbonate

.12 moles Na(CO3) 83.00 g Na(CO3) 1 liter 45 ml
1 liter
1 mole Na(CO3)
1000 ml

=.4482 grams of Na(CO3)

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14) 250 mL of 0.75 M lithium nitrite

.75 moles Li(NO2) 36.95 g Li(NO2) 1 liter 250 ml
1 liter
1 mole Li(NO2)
1000 ml

= 6.93 grams Li(NO2)

15) 56 mL of 1.1 M iron (II) phosphate
16) 6.7 L of 4.5 M ammonium nitrate
17) 4.5 mL of 0.05 M magnesium sulfate
18) 90 mL of 1.2 M BF3