Calculate molar concentration of 500 cm^3 glucose solution with mass part 5%

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New poster has been reminded to show their work on schoolwork problems
Homework Statement:
The question is:
Calculate molar concentration of 500 cm^3 glucose solution with mass part 5%. Density of the solution is 1g/cm^3.
Relevant Equations:
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Answers and Replies

  • #2
phyzguy
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We can't help until you show some effort. What have you learned about calculating molar concentrations? Have you solved similar problems in the past? What don't you understand about the problem?
 
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We can't help until you show some effort. What have you learned about calculating molar concentrations? Have you solved similar problems in the past? What don't you understand about the problem?
I found the solution mass which is 500 g. Then I found the glucose mass (25 g). And now I don’t know how to continue. I know I have to find the M(C6H12O6) in order to find the amount of the solute and then the concentration. This is where I am stuck.
 
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We can't help until you show some effort. What have you learned about calculating molar concentrations? Have you solved similar problems in the past? What don't you understand about the problem?
The formulas I use are:
m(solution) = V x p
m(C6H12O6) = w(solution) x m (solution)
And this is where I struggle:
n(C6H12O6) = m(C6H12O6)/M(C6H12O6)
c = n/V
 
  • #5
phyzguy
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What is the weight of one mole of glucose? So how many moles is 25g of glucose?
 
  • #6
Borek
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I found the solution mass which is 500 g.

Beware: that's assuming density of the solution is 1 g/mL. The more concentrated the solution, the less likely it is to be true. The only way to be sure is to check the density of the solution in density tables - you can google them, or - if the question is from the course - they were probably handed to you or are in a recommended book.

For 5% glucose solution error you are making assuming its density is 1 g/mL is a little below 2%.

How do you calculate molar mass of a substance knowing its formula?
 
  • #7
phyzguy
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@Borek, the problem gives the density at 1 g/cm^3.
 
  • #8
Borek
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the problem gives the density at 1 g/cm^3.
Ah, good point. Still, worth of noting it is only an approximation.
 
  • #9
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The formulas I use are:
m(solution) = V x p
m(C6H12O6) = w(solution) x m (solution)
And this is where I struggle:
n(C6H12O6) = m(C6H12O6)/M(C6H12O6)
c = n/V
Are you really saying that you know the chemical formula for a substance and you don’t know how to calculate its molecular weight?
 

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