Answer:
At least four such rinses.
Explanation:
<h3>What's the concentration after each rinse?</h3>
Let denotes the concentration of the residue after the th rinse. can be any non-negative integer (which includes zero.)
The initial concentration of the solution in the flask is . That is:
.
of this solution is left in the flask after it is emptied for the first time.
The solvent will increase the volume of the solution from to . At this point the number of moles of solute in the flask stays unchanged. The concentration of the residue will drop to of the initial value. That is:
.
Repeat this process, and the concentration of the residue will drop by a factor of again. That is:
.
Summarize these values in a table:
.
The trend in is similar to that of a geometric series with
- The concentration of the residue before any rinse, , and
- Common ratio .
Again, refer to the trend in as the value of increases. The general formula for , the concentration after the th rinse, will be:
.
As a side note, . As a result, the value of will decrease but stay positive as the value of increases. Increasing the number of rinses will indeed reduce the concentration of the residue.
<h3>How many such rinses are required? </h3>
In other words, what's the minimum value of for which ?
Recall that
.
As a result, should satisfy the condition:
.
Multiply both sides by
,
which is positive and will not change the direction of the inequality:
.
Take the natural logarithm of both sides of the inequality. The function is increasing as increases on the range . This function will not change the direction of the inequality, either.
.
Apply the power rule of logarithms: is an exponent inside the logarithm. That will be equivalent to an expression where is a coefficient in front of the logarithm operator:
.
.
Multiply both sides by
.
Keep in mind that logarithms of numbers between 0 and 1 (excluding 0 and 1) are negative. The reciprocal of a negative number is still negative. The direction of the inequality will change. That is:
.
The right-hand side is approximately 3.95. However, has to be an integer. The smallest integer that is greater than 3.95 is 4. In other words, it takes at least four such rinses to reduce the residual concentration to under 0.000100 M (0.000090 M to be precise.) Three such rinses will give only 0.00900 M.