Questions: A geochemist measures the concentration of salt dissolved in Lake Parsons and finds a concentration of 19.59 g · L^(-1). The geochemist also measures the concentration of salt in several nearby non-isolated lakes, and finds an average concentration of 2.23 g · L^(-1). Assuming the salt concentration in Lake Parsons before it became isolated was equal to the average salt concentration in nearby non-isolated lakes, calculate the percentage of Lake Parsons which has evaporated since it became isolated. Be sure your answer has the correct number of significant digits.

Solution Steps

We need to calculate the percentage of Lake Parsons that has evaporated since it became isolated. Initially, the salt concentration was the same as the nearby lakes, but now it is higher due to evaporation.

• Initial concentration of salt in Lake Parsons: \( C_i = 2.23 \, \text{g/L} \)
• Final concentration of salt in Lake Parsons: \( C_f = 19.59 \, \text{g/L} \)

The amount of salt in the lake remains constant before and after evaporation. Therefore, the initial volume \( V_i \) and final volume \( V_f \) are related by: \[ C_i \times V_i = C_f \times V_f \]

Rearrange the equation to solve for \( V_f \): \[ V_f = \frac{C_i \times V_i}{C_f} \]

The percentage of the lake that has evaporated is given by: \[ \text{Percentage Evaporated} = \left(1 - \frac{V_f}{V_i}\right) \times 100\% \]

Substitute \( V_f \) from Step 4: \[ \text{Percentage Evaporated} = \left(1 - \frac{C_i}{C_f}\right) \times 100\% \]

Substitute the given concentrations: \[ \text{Percentage Evaporated} = \left(1 - \frac{2.23}{19.59}\right) \times 100\% \]

Calculate the fraction: \[ \frac{2.23}{19.59} \approx 0.1138 \]

Calculate the percentage: \[ \text{Percentage Evaporated} = (1 - 0.1138) \times 100\% \approx 88.62\% \]

Final Answer