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## Q. 4.1

The average concentration of chloride ion in seawater is $19.353 g Cl^{–}/kg$ solution. The World Health Organization (WHO) recommends that the concentration of $Cl^{–}$ ions in drinking water not exceed 250 ppm $(2.50 × 10² ppm)$. How many times as much chloride ion is there in seawater than in the maximum concentration allowed in drinking water?

## Verified Solution

Collect and Organize We are given the concentrations of $Cl^{–}$ ions in two different units, $g Cl^{–}/kg$and ppm, and we are asked to determine the ratio of $Cl^{–}$ ions in seawater to the upper limit of $Cl^{–}$ ions allowed in drinking water.

Analyze We know how to create conversion factors from equalities. We can convert the seawater concentration to milligrams of $Cl^{–}$ per kilogram of seawater by using the conversion factor 10³ mg/1 g and then use the fact that 1 mg solute/kg solution = 1 ppm. We expect the value in mg/kg to be much larger than the value given in g/kg.

Solve

$19.353 \frac{\sout{g} Cl^{–}}{kg seawater} \times \frac{10^{3} mg}{\sout{g}} =19,353 \frac{mg Cl^{–}}{kg seawater} =19,353 ppm Cl^{–}$

Next we take the ratio of the two concentrations:

$\frac{19,353 ppm Cl^{–} in seawater}{250 ppm Cl^{–} in drinking water}=77.4$

The concentration of $Cl^{–}$ ions in seawater is 77.4 times greater than would be acceptable in drinking water.

Think About It Common sense says that seawater is much saltier than drinking water.
Drinking seawater induces nausea and vomiting, and it may even cause death. We could also have converted the drinking water concentration to grams of $Cl^{–}$ per kilogram of drinking water and compared that number with $19.353 g Cl^{–}/kg seawater$. It does not matter which units you choose when making a comparison as long as the units of the two numbers are the same.