Question

1. you have a solution mixed at a concentration of 5 g/ml and need 100 gal mixed at 4 g/ml what do you do?
2. you wish to count the number of bacteria in 1 ml of a solution but there are too many to count! so you mix the 1 ml with 9 ml of water to dilute it, and then draw 1 ml of the new mixture. still too many, so you now mix with 9 ml of water and draw 1 ml of this new solution and now count 75 nasty little bacteria swimming around. how many bacteria are in the original solution?

Explanation:

Step1: Determine the dilution factor

The first dilution: \(1\) ml of solution + \(9\) ml of water. The total volume \(V_2 = 1 + 9=10\) ml. Using \(C_1V_1 = C_2V_2\), if \(V_1 = 1\) ml and \(V_2 = 10\) ml, the dilution factor \(D_1=\frac{V_2}{V_1}=10\).

The second dilution: \(1\) ml of the first - diluted solution + \(9\) ml of water. The total volume \(V'_2=1 + 9 = 10\) ml. The dilution factor \(D_2=\frac{V'_2}{V'_1}=10\) (where \(V'_1 = 1\) ml).

The total dilution factor \(D = D_1\times D_2=10\times10 = 100\).

Step2: Calculate the number of bacteria in the original solution

We know that in the second - diluted \(1\) ml of solution, the number of bacteria \(C_2 = 75\).

Using \(C_1V_1=C_2V_2\) (where \(V_1 = 1\) ml, \(V_2\) (total dilution volume factor) \(= 100\) (since the total dilution factor is \(100\) and we assume \(V_1 = 1\) ml of the original solution is equivalent to the diluted \(1\) ml after two - step dilution in terms of the formula application for counting).

\(C_1=\frac{C_2\times V_2}{V_1}\)

Since \(C_2 = 75\), \(V_2=100\), \(V_1 = 1\)

\(C_1=75\times100\)

Answer:

There are \(7500\) bacteria in the original \(1\) ml of solution.