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PROBLEM SETS: %BY MASS, % BY VOLUME, MOLARITY, MOLALITY (show your solution)
1. What is the molarity of a solution in which 0.850 grams of ammonium nitrate are dissolved in 345 mL of solution?
2. Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water.
3. Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C10H8) in 54.3 g benzene (C6H6).
4. What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?
5. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?
6. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?

Respuesta :

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The concentration of a substance can be expressed using molarity, molality or percent.

What is concentration?

The term concentration has to do with the amount of substance present in solution. Now let us solve the problems individually;

a) Number of moles =  0.850 grams/80 g/mol = 0.011 moles

molarity = 0.011 moles/345 * 10^-3 L = 0.032 M

b) Number of moles =  13.5g /58 g/mol = 0.23 moles

molality = 0.23 moles/250 * 10^-3 Kg = 0.92 m

c) Number of  moles = 16.5 g/128 g/mol = 0.13 moles

molality = 0.13 moles/54.3  * 10^-3 Kg =2.39 m

d) Total mass present =  12 g +  18 g + 25 g + 500 g = 555 g

mass percent of calcium sulfate =  12 g/555 g * 100/1 = 2.2 %

mass percent of sodium nitrate =  18 g/ 555 g  * 100/1 = 3.2 %

mass percent of potassium chloride = 25 g / 555 g  * 100/1 = 4.5%

mass percent of water =  500 g / 555 g  * 100/1 =90.1%

e) Total mass present = 125 g + 1500g = 1625 g

Mass percent of NaCl =  125 g/1625 g * 100/1 = 7.7%

f) Total volume of solution = 15 L +  28 L  = 43 L

percent by volume of acetone = 15 L/43 L * 100/1 =  34.9%

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