Homework Answers Molarity & Molality Worksheet 40.0 g NaOH

Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Concentration Problems: (Show your work for full credit!) 1) If you dissolve 2...

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Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Concentration Problems: (Show your work for full credit!) 1) If you dissolve 20.0 g of NaOH in 250.0 mL of water, what is the molarity of the solution? 20.0g NaOH

1 mol NaOH

= 0.500 mol

40.0 g NaOH

0.500 mol 0.250 L

= 2.00 mol/L

2) If you dissolve 350.0 g of K2CrO4 in 2.50 Kg of water, what is the molality of the solution? 350.0 g K2CrO4

1 mol K2CrO4

1.80 mol

= 1.80 mol

2.50 kg

194.2 g K2CrO4

= 0.720 mol/kg

3) If you dissolve 20.0 g of magnesium chloride (check the formula!) in 70.0 mL of water, what is the molarity of the solution? 20.0 g MgCl2

1 mol MgCl2

= 0.210 mol

95.3 g MgCl2

0.210 mol

= 3.00 mol/L

0.0700 L

4) If you dissolve 75.0 g of aluminum nitrate (check the formula!) in 750.0 g of water, what is the molality of the solution? 75.0 g Al(NO3)3

1 mol Al(NO3)3

= 0.352 mol

213.0 g Al(NO3)3

0.352 mol 0.7500 kg

= 0.469 mol/kg

5) If the molarity of a solution is 0.500 mol/L, and it is made from 250.0 mL of water, how many moles of solute were added to the solution? mol 0.2500 L

= 0.500 mol/L

Mol = 0.125 mol solute

6) If the molality of a CaCl2 solution is 0.100 mol/Kg, and was made from 500.0 mL of water, how many grams of CaCl2 were needed to make the solution? (Hint: How do you change mL to g for water?) mol 0.500 L 0.0500 mol CaCl2

= 0.100 mol/kg

111.1 g CaCl2 1 mol CaCl2

Mol = 0.0500 mol CaCl2

= 5.56 g CaCl2

7) If the molarity of an Iron (III) Chloride solution is 1.25 mol/L and was made from 100. mL of water, how many grams of Iron (III) Chloride were needed to make the solution? mol

= 1.25 mol/L

0.100 L 0.125 mol FeCl3

Mol = 0.125 mol FeCl3

162.4 g FeCl3

= 20.3 g FeCl3

1 mol FeCl3 8) If the molarity of a Ca3(PO4)2 solution is 0.750 mol/L and 100.0 grams of Ca3(PO4)2 were used to make the solution, how much water was needed to dissolve the Ca3(PO4)2 ? 100.0 g Ca3(PO4)2

1 mol Ca3(PO4)2

= 0.322 mol Ca3(PO4)2

310.3 g Ca3(PO4)2 0.322 mol Volume

= 0.750 mol/L

Volume = 0.429 L

9) If there are 4.00 ppm of fluoride ions in our water, what mass of water would you need to measure out to ensure that you have one fluoride ion? (Assume 3 significant digits for calculations)

1 million molecules H2O

1 mol H2O

18.0 g H2O

23

1 mol H2O

6.02x10 molecules H2O

-17

= 2.99 x 10

g H 2O

But this is the mass that 4 fluoride ions is found in, so you must divide this by 4 to find the mass -18 of water that contains one fluoride ion: 7.48 x 10 g Extra Credit If the acceptable amount of mercury in our drinking water is 0.00200 ppm, what would the molarity of this solution be? 0.002 atoms Hg

1 mol Hg 23

-27

= 3.32 x 10

mol Hg

6.02x10 atoms Hg 1 million molecules H2O

1 mol H2O

18.0 g H2O

23

1 mol H2O

6.02x10 molecules H2O

-27

3.32 x 10

-20

2.99 x 10

mol Hg L H 2O

=M

-17

= 2.99 x 10 g H2O (same for mL using density!)

-7

Molarity = 1.11x10 mol/L