Question

1. a mole is a strange unit of measurement giving the amount of a substance that contains 6.022×10²³ \elementary entities\ (atoms, molecules, etc.). the molarity is a concentration measured in moles per liter. you have a solution with a molarity of 10⁻³ moles/l and need 100 ml at 10⁻⁴ moles/l. what do you do?

m₁ v₁ = m₂ v₂

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?

c₁ v₁ = c₂ v₂

1. 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?

c₁ v₁ = c₂ v₂

Explanation:

Step1: Determine the dilution factor for each step

The first dilution: \(1\space ml\) of solution + \(9\space ml\) of water. The total volume is \(V = 1+9=10\space ml\). The dilution factor \(D_1=\frac{10}{1}\)

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

The overall dilution factor \(D = D_1\times D_2=\frac{10}{1}\times\frac{10}{1}=100\)

Step2: Calculate the number of bacteria in the original solution

We know that the number of bacteria in the final \(1 - ml\) sample \(N_{final}=75\)

Let \(N_{original}\) be the number of bacteria in the original \(1 - ml\) solution.

Using the relationship \(N_{original}=N_{final}\times D\)

Substitute \(N_{final} = 75\) and \(D = 100\) into the formula: \(N_{original}=75\times100\)

Answer:

\(7500\) bacteria are in the original solution.