Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: mrtinkles on November 05, 2020, 05:25:17 AM
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Hello, is anyone able to help on how to go about doing this?
An experiment requires a solution which is 160 mM glycine and 10 mM MgSO4. What you have available in the lab is distilled water, a 2.2 M stock solution of glycine, and a 0.5 M MgSO4 stock solution. The experiment requires 5 ml of the solution in a test tube.
(a) What volume of the stock glycine solution do you need to transfer to the test tube?
(b) What volume of stock MgSO4 solution do you need to transfer to the test tube?
(c) What volume of distilled water do you need to transfer to the test tube?
(d) How many moles of glycine are in the test tube?
(e) How many mg of glycine does this represent? (The mass of 1 mole of Glycine is 75g.)
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Where are the problems? Show your work First.
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The problem is I dont know what I'm meant to be doing with it so unsure where to start. It would be nice if someone could explain how to do the first one so I can work from there
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The problem is with diluting the solutions. Every general chemistry textbook does this.
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The problem is I dont know what I'm meant to be doing with it so unsure where to start. It would be nice if someone could explain how to do the first one so I can work from there
Start first you have to do a 5 ml solution. Calculate the mole of Glycine and magnesiumsulfate with the given target values.
If you have this how much ml you need by using the stocksolutions. The last is to calculate the water by subtraction of all volume from 5 ml.
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How does this look?
An experiment requires a solution which is 160 mM glycine and 10 mM MgSO4. What you have available in the lab is distilled water, a 2.2 M stock solution of glycine, and a 0.5 M MgSO4 stock solution. The experiment requires 5 ml of the solution in a test tube.
(a) What volume of the stock glycine solution do you need to transfer to the test tube?
160mM / 1000 = 0.16M
C1V1 = C2V2
= 2.2M x V1 = 5ml x 0.16M / 2.2M
= v1 = 5ml x 0.16M / 2.2M
= 0.36ml
V1 - V2
= 5ml - 0.36ml = 4.64ml water needed (without other components)
?
Thanks
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You forgot about magnesium sulfate.
Correctly, you add water to the final volume. In the case of relatively dilute solutions, you may assume the additivity of volumes.
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Okay thanks, so the calculations from a - c should hopefully look like this?
An experiment requires a solution which is 160 mM glycine and 10 mM MgSO4. What you have available in the lab is distilled water, a 2.2 M stock solution of glycine, and a 0.5 M MgSO4 stock solution. The experiment requires 5 ml of the solution in a test tube.
(a) What volume of the stock glycine solution do you need to transfer to the test tube?
160mM / 1000 = 0.16M glycine
C1V1 = C2V2
= 2.2M x V1 = 5ml x 0.16M / 2.2M
= V1 = 5ml x 0.16M / 2.2M
= 0.36ml of glycine
(b) What volume of stock MgSO4 solution do you need to transfer to the test tube?
10mM / 1000 = 0.01M MgSO4
C1V1 = C2V2
= 0.5M x V1 = 5ml x 0.01M / 0.5M
= V1 = 5ml x 0.01M / 0.5M
= 0.1ml of MgSO4
(c) What volume of distilled water do you need to transfer to the test tube?
V1 - V2
= 5ml - 0.46ml
= 4.54ml water needed
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Yes correct. But as already mentioned add water to 5 ml.
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I dont really understand it but if I needed 0.46 ml of the chemicals then would 4.54ml of water not be right to make it up to 5ml?
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During mixing volume contraction or expansion can happen volume of solutions can not added. This is only valid for water.
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Oh thanks I get it now.
I'm trying to work out the next two questions now.
(d) How many moles of glycine are in the test tube?
I suppose this is using the information I just worked out but not sure in which way?
(e) How many mg of glycine does this represent? (The mass of 1 mole of Glycine is 75g.)
Is this GFM x moles x 1000
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The molarity is given, also the volume. You got the result also between the first calculation.