Question

you ask your lab partner to add 1mL to a 10mL water sample to 99 mL of diluent. Your partner inadvertently adds the entire sample of water to the diluent bottle. What was the intended dilution factor and what is the actual dilution factor?

Answer 1

The intended dilution factor => 1 mL in a total volume of 100 (=99+1) mL = 1/100 or 10^-2.

The actual dilution factor => 10 mL in a total volume of 109 (=99+10) mL = 10/109 = 0.0917431193 or 9.174 x 10^-2​​​​​​.

Answer 2

The intended dilution factor is the ratio of the volume of the original sample to the total volume after dilution. In this case, the intended dilution factor was supposed to be:

Intended Dilution Factor = Volume of Original Sample / Total Volume after Dilution Intended Dilution Factor = 1 mL / (1 mL + 99 mL) Intended Dilution Factor = 1 mL / 100 mL Intended Dilution Factor = 1/100

So, the intended dilution factor was 1/100.

However, due to the mistake made by the lab partner, the entire 1 mL of the original sample was added to the diluent bottle, resulting in a different total volume than intended. The actual dilution factor is the ratio of the volume of the original sample to the total volume after the mistake was made. Let's calculate the actual dilution factor:

Actual Dilution Factor = Volume of Original Sample / Total Volume after the Mistake Actual Dilution Factor = 1 mL / (1 mL + 99 mL) Actual Dilution Factor = 1 mL / 100 mL Actual Dilution Factor = 1/100

So, the actual dilution factor is also 1/100, despite the mistake made during the dilution process.