Take stock Allura red SOLIDS (~496g/mol in molar mass) and calculate the volume of the container you want to use for your own stock solution; the container they have has a mol/L count(M) of 9.476*10^(-4)M. For their case, if this was a 1L container, we would take that number in mol/L and multiply it by the 496 grams, divide by 1 mol and multiply by 1L which gets 0.4700096 grams. Therefore, for 9.476*10^(-4)M in a *1 liter* container, you need ~0.47 grams of Allura red solids; mix it all the way up with water and you get the same stock solution as them. Your institution may require or even give a different stock solution (M), so keep that in mind. Also keep in mind if you use a different volume, you will have to multiply by different amounts (e.g, 50mL would have to multiply 0.05L in the prior equation). Good luck and good chemistry. 👍
This was such a wonderful and instructional video, great job and thank you!
Thank you. I ask my students to watch this before their lab.
is it okay to touch the bottom of the beaker with the glass pipette when picking up stock dye? Will this cause air bubbles?
It's ok, provided you keep the pipette at an angle to the bottom of the beaker.
@@DrRichardMusgrave Thank you!
how did u make the stock solution?
Take stock Allura red SOLIDS (~496g/mol in molar mass) and calculate the volume of the container you want to use for your own stock solution; the container they have has a mol/L count(M) of 9.476*10^(-4)M.
For their case, if this was a 1L container, we would take that number in mol/L and multiply it by the 496 grams, divide by 1 mol and multiply by 1L which gets 0.4700096 grams. Therefore, for 9.476*10^(-4)M in a *1 liter* container, you need ~0.47 grams of Allura red solids; mix it all the way up with water and you get the same stock solution as them.
Your institution may require or even give a different stock solution (M), so keep that in mind.
Also keep in mind if you use a different volume, you will have to multiply by different amounts (e.g, 50mL would have to multiply 0.05L in the prior equation).
Good luck and good chemistry. 👍