Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: ydogyy1 on September 19, 2009, 03:03:18 PM
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Hello, I have a question that is related to % w/v. The question is as follows:
You have a 20% w/v stock solution of a sugar of molecular weight 160 Daltons. How much of this stock solution will you need to use to...
a. prepare 10 ml of solution at 350 mg sugar per ml.
Attempt:
So 20% w/v stock solution = 0.2 g/mL
350 mg/mL x 10mL gives us 3500mg of the sugar, or 3.5g
so the amount of solution we want is
x mL = 3.5 g / (0.2g / mL) = 17.5mL ( is this correct?)
b. 7 ml of solution containing 15 micromoles of sugar
Attempt:
15umol = 1.5 x 10^-5 mol
1.5 x 10^-5 mol (160 Daltons / 1mol) = 2.4 x 10^-3 g
(2.4 x 10^-3g) / 7mL = 4.89 x 10^-5 g/mL
and then I am stuck. How are you supposed to make 4.89 x 10^-5g/mL solution from a 0.2g/mL stock solution?! Is it possible? it does say in the question If the specific solution cannot be made, explain why not.
thank you in advance!
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Hello, I have a question that is related to % w/v. The question is as follows:
You have a 20% w/v stock solution of a sugar of molecular weight 160 Daltons. How much of this stock solution will you need to use to...
a. prepare 10 ml of solution at 350 mg sugar per ml.
Attempt:
So 20% w/v stock solution = 0.2 g/mL
350 mg/mL x 10mL gives us 3500mg of the sugar, or 3.5g
so the amount of solution we want is
x mL = 3.5 g / (0.2g / mL) = 17.5mL ( is this correct?)
So to make 10ml of solution you use 17.5ml do that sound possible?
b. 7 ml of solution containing 15 micromoles of sugar
Attempt:
15umol = 1.5 x 10^-5 mol
1.5 x 10^-5 mol (160 Daltons / 1mol) = 2.4 x 10^-3 g
(2.4 x 10^-3g) / 7mL = 4.89 x 10^-5 g/mL
Correct except for the last part of this calculation,
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So to make 10ml of solution you use 17.5ml do that sound possible?
no, I suppose not :-[. could you give me any hints to nudge me in the right direction?
Correct except for the last part of this calculation,
o right, it should be 3.42 x 10^-4 g/mL. so with this concentration...how can we figure out how much of the stock solution to add?