The following modules will allow you to practice converting between units and creating solutions. These are skills that are super useful in any lab setting, so the overall goal of this exercise is to help you feel comfortable when doing this type of lab math.

Each module will have an explanation and a series of questions. To check your answers, click the "Show Answers" button. If you have an incorrect answer and want to try again, no problem! Simply re-enter your new answer and it will automatically be checked. If you don't want to see the feedback immediately, re-click the "Show Answers" button to toggle off the feedback.

This module is based on the Web Exercise template created by the #PsyTeachR team at the University of Glasgow, based on ideas from Software Carpentry.

Getting comfortable with units

Volumes

This section will help you practice converting between units of volume. Milliliters are abbreviated as 'mL' and microliters as 'uL'. Milliliters and microliters are volumes with a base in liters: one milliliter is one thousandth of a liter. The prefix 'milli' means 'thousandth'. Therefore, 5 milliliters is equivalent to 5/1,000 L or 0.005 L. The prefix micro means 'millionth'; Therefore, 5 microliters is equivalent to 5/1,000,000 or 0.000005 L.

Now let's practice these conversions!

  • 100 mL is L

  • 47 mL is L

  • 208 mL is L

  • 0.07 L is mL

  • 0.009 L is mL

  • 0.001 L is mL

  • 1 L is equivalent to

Because milliliters and microliters are on the same 'liters' base scale, this means they are related to one another. One milliliter is equivalent to one thousand microliters; conversely, one thousand microliters are equivalent to one milliliter. Therefore, 5 microliters are 5/1000 or 0.005 mL.

Let's practice these conversions!

  • 94 ul is mL

  • 207 uL is mL

  • 597 uL is mL

  • 0.013 mL is uL

  • 0.004 mL is uL

  • 0.001 mL is uL

  • 1000 uL are equivalent to

Concentrations

A concentration is a description of the amount of a substance within a volume. In biology and chemistry, concentrations are reported as molarities (abbreviated M), or the number of moles of a substance per liter of solvent. Therefore, 1M is equivalent to 1 mole of sbustance in 1 L of solvent.

Let's practice!

  • 100 M is moles of substance in 1 L of solvent.

  • 24 M is moles of substance in 1 L of solvent.

  • 46 M is moles of substance in 1 L of solvent.

  • 6 M is moles of substance in 500 mL of solvent.

  • 14 M is moles of substance in 500 mL of solvent.

  • 8 M is moles of substance in 100 mL of solvent.

When working in a lab, the concentrations are usually in the micromolar range. Recall that the prefix 'micro' means millionth. Because Molarity is abbreviated as 'M', micromolar as 'uM'. A molarity of 1 is equivalent to 1,000,000 uM. On the other hand, 1000 uM is equivalent to 0.001 (or 1000/1,000,000) M. For these questions, you'll also have to convert between uM and mM as well as mM and M, so be sure to pay attention to the units.

Let's practice these conversions! Note: If your answer is 1000 or above, there is no need to include commas (i.e 9999 instead of 9,999).

  • 1000 uM is M

  • 650 uM is mM

  • 9800 uM is M

  • 246 uM is mM

  • 35 uM is mM

  • 0.01 M is uM

  • 0.003 M is uM

  • 0.00942 M is uM

  • 0.206 M is mM

  • 0.18 M is mM

  • 500 uM is equivalent to

Combine your knowledge from the volumes and concentrations sections to answer the questions below:

  • 500 uM is moles of substance in 1 L of solvent.

  • 75 uM is moles of substance in 100 mL of solvent.

  • 27 M is moles of substance in 100 mL of solvent.

Making solutions

explanation explanation C1V1 = C2V2 etc

To solve for V1, divide C2*V2 by C1. In other words:

\[\mathbf{V1} = \frac{C2*V2}{C1}\]

Your lab manager asks you to make 500mL of a 500uM solution. You have a stock solution that is 1mM and a jug of deionized water.

Before solving for V1, make sure your units for C1 and C2 are the same. Notice that the C1 is in mM, so let’s convert C2 to mM as well

  • 500 uM is equivalent to

  • C1 =

  • C2 =

  • V1 =

  • V2 =

Now solve for V1:

  • V1 is mL

To create 500mL of a 500uM solution, add of the 1mM solution to of water.

Use the same approach to try the following problem:

Your lab manager asks you to make 100mL of a 100uM solution. You have a stock solution that is 250uM and a jug of deionized water.

  • C1 = uM

  • C2 = uM

  • V2 = mL

Now do the calculations to make the solution:

To create 100mL of a 100uM solution, add mL of the 250uM solution to mL of water.

Your lab manager asks you to make 15mL of a 0.5uM solution. You have a stock solution that is 100uM and a jug of deionized water.

To create 15mL of a 0.5uM solution, add uL of the 100uM solution to mL of water.

Your lab manager asks you to make 340mL of a 32uM solution. You have a stock solution that is 500uM and a jug of deionized water.

To create 340 mL of a 32 uM solution, add mL of the 500uM solution to mL of water.

Your lab manager asks you to make 13mL of a 1.9uM solution. You have a stock solution that is 100uM and a jug of deionized water.

To create 13 mL of a 1.9 uM solution, add uL of the 100uM solution to mL of water.

Your lab manager asks you to make 10 mL of a 25uM solution. You have a stock solution that is 1M and a jug of deionized water.

To create 10 mL of a 25 uM solution, add uL of the 1 uM solution to mL of water.

This solution is problematic because

A solution to this conundrum is to

This could be achieved by