Chapter 1, Problem 110AP

Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1 ppm chlorine, or 1 g of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 6.0 percent chlorine by mass and there are $2.0 \times {10}^4$ gallons (gal) of water in the pool $\left(\text{1 gal = 3}\text{.79 L; density of liquids = 1}\text{.0 g/mL}\right)$.

Interpretation Introduction

Interpretation:

The volume of a chlorine solution added to a for a given amount of water in the swimming pool is to be calculated.

Concept introduction:

Dimensional analysis is a way to convert units of measurement. In order to convert one unit to another, the relationship between those units is to be predicted. These relationships are called conversion factors. Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.

The conversion factor is a fraction obtained from a relationship between the units. It is written as a ratio and can be inverted to give two conversion factors.

$\text{1gal}=\text{3}\text{.79 L}$ the conversion factor is $\frac{\text{3}\text{.79\#x2009;L}}{\text{1gal}}$.

$\text{1 L}=\text{1000 mL}$ the conversion factor is $\frac{\text{1000\#x2009;m\#x2009;L}}{\text{1L}}$.

Answer to Problem 110AP

Solution: $1.3\times {10}^3\text{ mL}$

Explanation of Solution

Given information: The mass percent of chlorine in solution is $6.0\%$.

The water in the swimming pool is $2.0\times {10}^4\text{gal}$.

The density of liquids is $\text{1}\text{.0}\text{g}/\text{mL}$.

The accepted concentration of chlorine is 1 ppm.

The volume of water in the pool is $2.0\times {10}^4\text{gal}$.

The density of liquids is 1.0 g/mL.

The conversion factors are as follows:

$\text{1gal}=\text{3}\text{.79 L}$

$\text{1 L}=\text{1000 mL}$

Multiply $2.0\times {10}^4\text{gal}$ with the conversion factor $\frac{\text{3}\text{.79\#x2009;L}}{\text{1gal}}$

Volume = $\left(2.0\times {10}^4\text{gal}\right)\left(\frac{\text{3}\text{.79 L}}{\text{1gal}}\right)$

The volume of water in gallons is as follows:

$\left(2.0\times {10}^4\cancel{\text{gal}}\right)\left(\frac{\text{3}\text{.79 L}}{\text{1}\cancel{\text{gal}}}\right)=7.58\times {10}^4\text{L}$

Multiply the calculated volume of water in the pool with conversion factor $\frac{\text{1000\#x2009;m\#x2009;L}}{\text{1L}}$ and density $\frac{\text{1}\text{.0\#x2009;g\#x2009;}}{\text{1}\text{.0\#x2009;mL}}$ to get the mass of water in the pool.

The mass of water is as follows:

$\left(7.58\times {10}^4\cancel{\text{L}}\right)\left(\frac{\text{1000 }\cancel{\text{m L}}}{\text{1}\cancel{\text{L}}}\right)\left(\frac{\text{1}\text{.0 g }}{\text{1}\text{.0 }\cancel{\text{mL}}}\right)=7.58\times {10}^7\text{g}$

The accepted concentration of chlorine to be added to the pool is 1 ppm or1 g chlorine per million grams of water.

It can be expressed as follows:

$\text{1 g Cl}={\text{10}}^{\text{6}}{\text{g H}}_{\text{2}}\text{O}$

The conversion factor to obtain the mass of chlorine to be added in a given amount of water is $\frac{\text{1\#x2009;gCl}}{{\text{10}}^{\text{6}}{\text{\#x2009;g\#x2009;H}}_{\text{2}}\text{O}}$.

Multiply $7.58\times {10}^7\text{g}$ of water with $\frac{\text{1\#x2009;gCl}}{{\text{10}}^{\text{6}}{\text{\#x2009;g\#x2009;H}}_{\text{2}}\text{O}}$ to get the mass of chlorine to be added to water.

The mass of chlorine added is as follows:

$\left(7.58\times {10}^7\cancel{{\text{g H}}_{\text{2}}\text{O}}\right)\left(\frac{\text{1 gCl}}{{\text{10}}^{\text{6}}\cancel{{\text{g H}}_{\text{2}}\text{O}}}\right)=75.8\text{gCl}$

Mass percent of chlorine in the chlorine solution is 6.0 percent or 6.0 g Cl in 100 g H2O. Thus, multiply the calculated mass of chlorine with the conversion factor $\frac{{\text{100\#x2009;g\#x2009;H}}_{\text{2}}\text{O}}{\text{6}\text{.0\#x2009;g\#x2009;Cl}}$, and then with $\frac{\text{1}{\text{.0\#x2009;mLH}}_{\text{2}}\text{O}}{\text{1}{\text{.0\#x2009;gH}}_{\text{2}}\text{O\#x2009;}}$, to find the volume of the chlorine solution that needs to be added to the pool.

The volume of chlorine to be added in the pool is as follows:

$\left(75.8\cancel{\text{g}}\text{Cl}\right)\left(\frac{\text{100 }\cancel{\text{g}}{\text{ H}}_{\text{2}}\text{O}}{\text{6}\text{.0 }\cancel{\text{g }}\text{Cl}}\right)\left(\frac{\text{1}{\text{.0 mLH}}_{\text{2}}\text{O }}{\text{1}\text{.0 }\cancel{\text{g}}{\text{H}}_{\text{2}}\text{O}}\right)=1263.33{\text{mL H}}_{\text{2}}\text{O}$

The volume of chlorine is rounded-off to two significant figures that is $1.3\times {10}^3\text{ mL}$.

Conclusion

The volume of the chlorine solution to be added to a swimming pool is $1.3\times {10}^3\text{ mL}$.