Difference between volatile and semivolatile hydrocarbons compounds
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The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of H2SO4H2SO4 per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
EXAMPLE 13.4.213.4.2: MOLARITY AND MASS
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
What is the molarity of the solution?
What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
Use the concentration of the solute in parts per million to calculate the molarity.
Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
molarity=moleslitersolution=(12.7mg)(1g1000mg)(1mol78.114g)1.00Lmolarity=moleslitersolution=(12.7mg)(1g1000mg)(1mol78.114g)1.00L
=1.63×10−4M=1.63×10−4M
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
massofbenzene=(12.7mgbenzene)(250mL)1000mLmassofbenzene=(12.7mgbenzene)(250mL)1000mL
=3.18mg=3.18×10−3gbenzene=3.18mg=3.18×10−3gbenzene
EXERCISE 13.4.213.4.2: MOLARITY OF LEAD SOLUTION
The maximum allowable concentration of lead in drinking water is 9.0 ppb.
EXAMPLE 13.4.213.4.2: MOLARITY AND MASS
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
What is the molarity of the solution?
What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
Use the concentration of the solute in parts per million to calculate the molarity.
Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
molarity=moleslitersolution=(12.7mg)(1g1000mg)(1mol78.114g)1.00Lmolarity=moleslitersolution=(12.7mg)(1g1000mg)(1mol78.114g)1.00L
=1.63×10−4M=1.63×10−4M
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
massofbenzene=(12.7mgbenzene)(250mL)1000mLmassofbenzene=(12.7mgbenzene)(250mL)1000mL
=3.18mg=3.18×10−3gbenzene=3.18mg=3.18×10−3gbenzene
EXERCISE 13.4.213.4.2: MOLARITY OF LEAD SOLUTION
The maximum allowable concentration of lead in drinking water is 9.0 ppb.
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