1. how many mL of a 45.0% (m/v) solution of NaCl would you need in order to have

ID: 829628 • Letter: 1

Question

1. how many mL of a 45.0% (m/v) solution of NaCl would you need in order to have 12.0 g of NaCl?

2. A bottle of wine contains 750.0 mL of wine. If the wine is 12.00% (v/v) alcohol, how many mL of alcohol are in the bottle of wine?

3. A mixture of alcohol and water is 28.00% (v/v) alcohol. How many L of the mixture do I need in order to have 250.0 mL of alcohol?

4. A mixture of salt and water is 28.0 g salt and 60.0 g water. What is the % (m/m) salt?

5. A salt water mixture is 33.0% (w/w) salt. If there are 300.0 g salt in the mixture, what is the total weight of the mixture?

6. A sand-water mixture weighs 38.6 g. If the mixture is 15.0% (m/m) water, how many grams of water are in the mixture?

(1) %m/V = mass of NaCl/volume of solution x 100%

45.0 = 12.0/volume of solution x 100%

Volume of solution = 26.7 mL

(2) %v/v = volume of alcohol/volume of wine x 100%

12.00 = volume of alcohol/750.0 x 100%

Volume of alcohol = 90.00 mL

(3) %v/v = volume of alcohol/volume of mixture x 100%

28.00 = 250.0/volume of mixture x 100%

Volume of mixture = 892.9 mL = 0.8929 L

(4) %m/m = mass of salt/(mass of salt + mass of water) x 100%

= 28.0/(28.0 + 60.0) x 100% = 31.8%

(5) %w/w = weight of salt/weight of mixture x 100%

33.0 = 300.0/weight of mixture x 100%

Weight of mixture = 909.1 g = 909 g

(6) %m/m = mass of water/mass of mixture x 100%

15.0 = mass of water/38.6 x 100%

Mass of water = 5.79 g

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