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
