Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: apatel81093 on November 08, 2008, 07:05:07 PM
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A student placed 18.0 g of glucose ( C_6 H_12O_6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100.mL of the final solution?
please help
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We are here to help you, not do your work for you! (http://www.chemicalforums.com/index.php?topic=9037.0)
Show your attempts first!
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i actually did make an attempt
i got an answer of 1.8 grams but the computer said it was wrong
after that, i made another attempt but the answer i got was 90 g, which is way too high
now can u assist me please?
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Write out what you did, it will be easier to see your mistakes then.
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honestly, this is no help at all
ive tried writing the problem out over and over on multiple pieces of paper but i have no idea what is going on.
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How many moles of glucose is in your starting solution? How many moles are there after dilution? So, using the MW of glucose, with that many moles, how many grams do you have?
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im sry i don't understand
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You say you've done attempts, show us what you did (the math and explain what you know).
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How many moles of glucose is in your starting solution? How many moles are there after dilution? So, using the MW of glucose, with that many moles, how many grams do you have?
Sorry enahs, but it doesn't make sense to go through moles here. You start with mass, you end with mass, you should go using mass in the middle. Moles will only add to the confusion.
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1.8g is the correct answer.
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1.8g is the correct answer.
Uh oh, it is not :)
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How many moles of glucose is in your starting solution? How many moles are there after dilution? So, using the MW of glucose, with that many moles, how many grams do you have?
Sorry enahs, but it doesn't make sense to go through moles here. You start with mass, you end with mass, you should go using mass in the middle. Moles will only add to the confusion.
I disagree that it will be more confusing. After you make the first solution, and then you take 20mL of that. So now instead of doing the simple dilution calculations and using molecular weight, as the teacher and book should have shown numerous times, if you use mass you are assuming he/she is good with percentages, and then again in the end. Which they really should be good with percentages though. But I would rather assume they have been shown dilution calculations instead of percentages in a Chemistry class!
But, along with how many moles in your starting solution, I was intending to imply concentration.
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1.8g is the correct answer.
Uh oh, it is not :)
Yes I'm an idiot. 1/5th of 1/5th is 1/25th.
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Don't worry, you were just trying to show us the futility of simply giving an answer, with no working, weren't you?
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Don't worry, you were just trying to show us the futility of simply giving an answer, with no working, weren't you?
Yeh sure that was it.
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But I would rather assume they have been shown dilution calculations instead of percentages in a Chemistry class!
It is not even about dilution.
You dissolve some mass in some volume, you take part of that volume - this is simple fraction.
You take this fraction, in the next step you take other fraction of it.
This is simple kindergarten ;) math, nothing about chemistry or dilutions here, you don't even need to know what word "concentration" means ;D
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But I would rather assume they have been shown dilution calculations instead of percentages in a Chemistry class!
It is not even about dilution.
You dissolve some mass in some volume, you take part of that volume - this is simple fraction.
You take this fraction, in the next step you take other fraction of it.
This is simple kindergarten ;) math, nothing about chemistry or dilutions here, you don't even need to know what word "concentration" means ;D
I totally agree. And I knew the answer this way before I was done reading the first post (well, a rough estimate in my head).
But I am betting the teachers goal is to show how to use the concentration and dilution formulas. I know the original poster has been taught those; so I was just trying to nudge the original poster into using what they already know!
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Hey, we all "win some" and "lose some" in the old trick of trying to give hints that are useful, that a person can understand, if they're actually trying. We'll get lucking in the ol' "giving of hints and not answers" next time.
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hey guys im the guy that originally put up the question
i really need to know what the answer is.
some guy said it was 1.8g but i put that in and it was wrong
later, that same guy said it wasn't 1.8g but something else
please please please give me the answer :'(
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You need to show some work. You say you got 90g, how did you get that answer? What was the math that you did? You also say you got 1.8g, show us the math again! It is difficult to help you unless we know what you tried to do, if we see the math you did, then it will be easier to find your mistake.
We don't just give out answers, so stop asking.
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please please please give me the answer
As macman wrote - show your math that lead to 1.8 g and we will try to help you find what you did wrong - and why.
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C6H12O6 = 180.03 g/mol
18gC6H12O6 x 1mol/180.03 g = 0.099983 mol C6H12O6
0.099983 mol C6H12O6/0.1L = 0.9998 M
Since the diluted glucose must have the same number of moles as the original sample, then M1V1=M2V2, M is the Molarity of the sample and V its volume.
M2=(0.9998M)(0.02L)/(0.500L)=0.03999 M C6H12O6
0.03999 M C6H12O6X0.1L = 0.003999 mol C6H12O6
0.003999 mol C6H12O6 x 180.03g/mol =
.72
I don’t know if this is right
Like all of you guys said, I made an attempt and proved it
Now can u please help me?
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0.72g is a correct answer.
Note, that whole moles/molar masses/dilution work is not necessary here. There is much easier and much faster way of dealing with the problem.
You start with 18 g and dissolved it so that you have 100 mL of the solution. You then take 20 mL, which is precisely 1/5thof the solution - and it must contain 1/5th of the initial mass.
Then you dilute it to 500 mL and you take 100 mL - again 1/5th. So your final solution contains 1/5th of 1/5th - 1/25th of the initial mass of the glucose. 18g/25 = 0.72g.
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A student placed 18.0 g of glucose ( C_6 H_12O_6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100.mL of the final solution?
please help
what they are asking you to find is the mass in grams
i wont solve it its not that i cant its just that am not gona do your work lol
step1. calculate the moles use it to find mass easy cheese
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step1. calculate the moles use it to find mass easy cheese
Have you read the thread? Thre is absolutely no need to calculate moles, that's just a waste of time.