That Cytochrome C question in 109 workshop 2
Struggling with the Cytochrome C question in your 109 workshop 2.... Then this post is for you! Let me show you how I would do this question:
The Question:
A lot of information followed by...
The Biochemist prepared 2.5 ml of cytochrome c (Mr = 12000) in water (solution A). It was then diluted by addition of 1.0 ml solution A to 9.0 ml water to give solution B. Finally 2.0 ml solution B was diluted 20-fold (i.e. the ratio of the final volume to the initial volume is 20) with water to give a final solution C which was determined to have a concentration of 0.15 mg/ml.
Before I even look at the questions, I would draw out the information provided in a kind of flow diagram:
Ok, now I have a clear idea of what I have been told, I will look at the questions:
1) What volume of water was added to give the 20-fold dilution of cytochrome C in the preparation of solution C?
So, we know that we are talking about the second dilution step, where 2.0 ml of solution B was taken and diluted 20-fold. This means that the 2.0 ml of solution B was diluted 20x i.e. was put into a volume that was 20x bigger than the initial 2.0 ml. So that volume must be 2 x 20 = 40 ml. But watch out, 40 ml is the final volume after the dilution has been performed - if you did this in the lab you would take 2.0 ml of solution B and add it to 38 ml of water, so your final volume would be 40 ml.
So, the answer to this question is 38 ml.
We can add this information to our flow diagram:
2) What was the concentration of cytochrome C in the original solution A?
Well, we know that solution C has a concentration of 0.15 mg/ml, and we know that this concentration has been made after two dilution steps - the first step was a 10x dilution, and the second step was a 20x dilution. Personally, I would figure out the answer to this question in two steps:
- first I would calculate the concentration of solution B: solution B was diluted 20x to make solution C. This means that solution B must be 20x more concentrated than solution C, so solution B has a concentration of 0.15 x 20 = 3.0 mg/ml
- then I would calculate the concentration of solution A: solution A was diluted 10x to make solution B. This means that solution A must be 10x more concentrated than solution B, so solution A has a concentration of 3.0 x 10 = 30 mg/ml
I would add this information to my diagram:
3) How much cytochrome C was dissolved in the 2.5 ml water to prepare solution A?
Well, you now know that solution A has a concentration of 30 mg/ml. You also know that the biochemist made 2.5 ml of solution A. Therefore, seeing as there are 30 mg in every ml (30mg/ml) and 2.5 mls were made, that must mean that in these 2.5 mls there will be 30 x 2.5 = 75 mg
4) What was the molar concentration of solution B?
Well, there are various ways you could go about figuring this out - you could find the Molar concentration of solution A or C, and then do a dilution factor calculation to find out the Molar concentration of solution B (if you had found the Molar concentration of solution A, you would have to then divide this by 10 to get the concentration of solution B, if you had found the Molar concentration of solution C, you would have to then times this by 20 to get the concentration of solution B).
I think I would dive straight in with solution B itself:
So, to find a concentration, we need our trusty formula:
Now, to use this formula we need to know how many moles of cytochrome C are in solution B, and the volume of solution B.
Lets start with the easy one - the volume of solution B must be 10 ml - why? Because you are told that 1 ml of solution A was added to 9 ml of water to make solution B.
Now, how many moles of cytochrome C are there in solution B? Another equation is needed:
We know that Mr is 12000 (this was given to us in the question). But how many grams of cytochrome C are there in solution B? Well, we know that solution B has a concentration of 3 mg/ml (we figured this out as part of question 2) and we know that there are 10 ml of solution B. Therefore, if there are 3 mg of cytochrome C in every ml (i.e. 3.0 mg/ml) and we have 10 ml in total, then there must be 3.0 x 10 = 30 mg of cytochrome C in solution B.
Ok, so know we can calculate how many moles of cytochrome C there are in 10 ml of solution B:
Now we can go back to the concentration equation:
Phew, that was a long explanation!!
One last thing - I wrote all my answers with units and to 2 significant figures (the lowest accuracy given in the question was 2SF, so you have to stick with this in your answers)
Hope it helps!
The Question:
A lot of information followed by...
The Biochemist prepared 2.5 ml of cytochrome c (Mr = 12000) in water (solution A). It was then diluted by addition of 1.0 ml solution A to 9.0 ml water to give solution B. Finally 2.0 ml solution B was diluted 20-fold (i.e. the ratio of the final volume to the initial volume is 20) with water to give a final solution C which was determined to have a concentration of 0.15 mg/ml.
Before I even look at the questions, I would draw out the information provided in a kind of flow diagram:
Ok, now I have a clear idea of what I have been told, I will look at the questions:
1) What volume of water was added to give the 20-fold dilution of cytochrome C in the preparation of solution C?
So, we know that we are talking about the second dilution step, where 2.0 ml of solution B was taken and diluted 20-fold. This means that the 2.0 ml of solution B was diluted 20x i.e. was put into a volume that was 20x bigger than the initial 2.0 ml. So that volume must be 2 x 20 = 40 ml. But watch out, 40 ml is the final volume after the dilution has been performed - if you did this in the lab you would take 2.0 ml of solution B and add it to 38 ml of water, so your final volume would be 40 ml.
So, the answer to this question is 38 ml.
We can add this information to our flow diagram:
2) What was the concentration of cytochrome C in the original solution A?
Well, we know that solution C has a concentration of 0.15 mg/ml, and we know that this concentration has been made after two dilution steps - the first step was a 10x dilution, and the second step was a 20x dilution. Personally, I would figure out the answer to this question in two steps:
- first I would calculate the concentration of solution B: solution B was diluted 20x to make solution C. This means that solution B must be 20x more concentrated than solution C, so solution B has a concentration of 0.15 x 20 = 3.0 mg/ml
- then I would calculate the concentration of solution A: solution A was diluted 10x to make solution B. This means that solution A must be 10x more concentrated than solution B, so solution A has a concentration of 3.0 x 10 = 30 mg/ml
I would add this information to my diagram:
3) How much cytochrome C was dissolved in the 2.5 ml water to prepare solution A?
Well, you now know that solution A has a concentration of 30 mg/ml. You also know that the biochemist made 2.5 ml of solution A. Therefore, seeing as there are 30 mg in every ml (30mg/ml) and 2.5 mls were made, that must mean that in these 2.5 mls there will be 30 x 2.5 = 75 mg
4) What was the molar concentration of solution B?
Well, there are various ways you could go about figuring this out - you could find the Molar concentration of solution A or C, and then do a dilution factor calculation to find out the Molar concentration of solution B (if you had found the Molar concentration of solution A, you would have to then divide this by 10 to get the concentration of solution B, if you had found the Molar concentration of solution C, you would have to then times this by 20 to get the concentration of solution B).
I think I would dive straight in with solution B itself:
So, to find a concentration, we need our trusty formula:
Concentration = Moles/Volume
Now, to use this formula we need to know how many moles of cytochrome C are in solution B, and the volume of solution B.
Lets start with the easy one - the volume of solution B must be 10 ml - why? Because you are told that 1 ml of solution A was added to 9 ml of water to make solution B.
Now, how many moles of cytochrome C are there in solution B? Another equation is needed:
Moles = Mass/Mr
We know that Mr is 12000 (this was given to us in the question). But how many grams of cytochrome C are there in solution B? Well, we know that solution B has a concentration of 3 mg/ml (we figured this out as part of question 2) and we know that there are 10 ml of solution B. Therefore, if there are 3 mg of cytochrome C in every ml (i.e. 3.0 mg/ml) and we have 10 ml in total, then there must be 3.0 x 10 = 30 mg of cytochrome C in solution B.
Ok, so know we can calculate how many moles of cytochrome C there are in 10 ml of solution B:
Moles = Mass/Mr
Moles = 30mg/12000
Moles = 0.0025 mmoles
(note, my unit is mmoles since the mass was in mg)
Now we can go back to the concentration equation:
Concentration = Moles/Volume
Concentration = 0.0025mmoles/0.01L
(note, 10 mls has been converted to L)
Concentration = 0.25mM
(note the unit is mM since the moles were in mmoles)
Phew, that was a long explanation!!
One last thing - I wrote all my answers with units and to 2 significant figures (the lowest accuracy given in the question was 2SF, so you have to stick with this in your answers)
Hope it helps!
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