Question:
A geochemist measures the concentration of salt dissolved in Lake Parsons, an isolated salt lake. He finds a concentration of 74 gL−1. The geochemist also measures the concentration of salt in several nearby non-isolated lakes, and finds an average concentration of 6.5 gL−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.
Answer:
The percentage of Lake Parsons evaporated is 91 %
Explanation:
Let us say that Lake Parsons water content initially = X and salt content = Y
Therefore,
![(Y)/(X) = 6.5](https://img.qammunity.org/2021/formulas/chemistry/college/yr0shb2nigdlwkbl75wff5iet59sr1zlrz.png)
Y = 6.5 X
After Lake Parsons became isolated, its salt content remained unchanged i.e. Y
However, its water content decreased due to evaporation. Suppose now its water content = Z
Therefore we can write
![(Y)/(Z) = 74](https://img.qammunity.org/2021/formulas/chemistry/college/hk2eq1bb3px8eumno3yjct6wxlnysxxuez.png)
Substituting the Y value
\frac{6.5X}{Z} = 74
![(X)/(Z) = (74)/(6.5)](https://img.qammunity.org/2021/formulas/chemistry/college/4eycx9kb75bljxsd7bx87hf18mtgl43yye.png)
![(Z)/(X)= (6.5)/(74)](https://img.qammunity.org/2021/formulas/chemistry/college/dnupku8tvkpiiosp30ozhj4hp0lngk6eds.png)
![(Z)/(X) = 0.0878](https://img.qammunity.org/2021/formulas/chemistry/college/pu0ae74pg96avz0jn6b8yedrdnoxqd4fcj.png)
![(Z)/(X)= 8.78](https://img.qammunity.org/2021/formulas/chemistry/college/f93n0eu2zmv0sd9gon0ryo3hr068i6pecu.png)
Therefore the Lake Parsons have now 8.78 % of the initial water.
Now the percentage of Lake Parsons evaporated
= (100 - 8.78)
= 91.22
= 91 % (rounding to 2 significant digits)