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Topic: Dextrose solution Question  (Read 461 times)

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Offline m3thlab

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Dextrose solution Question
« on: December 14, 2019, 07:27:19 PM »
I have a homework question that I can not seem to figure out. I know it is simple but cannot seem to grasp it.
A 5% Dextrose solution is 0.5% Dextrose (g/mL). How many grams of Dextrose are needed for 1.7 L of solution?
The part that is seriously throwing me off is the 5% and .5%. I dont understand what it means that a 5% solution is .5% dextrose. If anyone could at least give me a pointer or indicate how I should go about solving a problem like this Id be extremely grateful!

Offline AWK

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Re: Dextrose solution Question
« Reply #1 on: December 15, 2019, 01:41:49 AM »
This is additional and unnecessary information. The task itself is formulated in a suspicious manner.
Should be:
How many grams of Dextrose is needed to make 1.7 L of 5% (m/v) solution?
« Last Edit: December 15, 2019, 04:26:37 AM by AWK »
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Offline Borek

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Re: Dextrose solution Question
« Reply #2 on: December 15, 2019, 04:31:31 AM »
There is a mistake in the question formulation, there is no way to say what it is really asking. No wonder you have problems solving it.

First phrase was probably intended to say something like "A 5% Dextrose solution is 0.05 (g/mL) Dextrose" or "A 50% Dextrose solution is 0.5 (g/mL) Dextrose."

I wouldn't be so sure AWK's guess is the right one. There are any other possible ones, like: how many grams of Dextrose are needed for making 1.7 L of 5% solution? or: how many grams of 50% Dextrose solution are needed for making 1.7 L of 5% solution? or even: how many grams of Dextrose are needed for making 1.7 L of 0.5% solution? It is quite likely that the target is 5% solution, as this is a common concentration used for IV administration, but the question doesn't state that.
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Offline AWK

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Re: Dextrose solution Question
« Reply #3 on: December 15, 2019, 05:35:37 AM »
Quote
how many grams of 50% Dextrose solution
At 25°C dextrose reaches 45 % concentration.
« Last Edit: December 15, 2019, 06:36:13 AM by AWK »
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Offline Borek

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Re: Dextrose solution Question
« Reply #4 on: December 15, 2019, 06:55:46 AM »
Apparently it depends on the source of the data and the 50% solubility is somewhere around the room temperature.

It still doesn't make your interpretation of what the question is unique, there are other possibilities. The only way to be sure is to contact the author of the question and ask for clarification.
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Offline MNIO

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Re: Dextrose solution Question
« Reply #5 on: December 26, 2019, 04:53:43 PM »
I suspect you have a typo.

instead of
  A 5% Dextrose solution is 0.5% Dextrose (g/mL).
it should be either
  A 0.5% dextrose solution is 0.5% dextrose (g/mL)
  A 5% dextrose solution is 5.0% dextrose (g/mL)

either way is a bit redundant.  A better way to approach this is
  0.5% dextrose = 0.5g dextrose / 100mL solution
likewise
  5% dextrose = 5g dextrose / 100mL solution

here's an example calc
    1.7L solution     10*100mL      0.5g dextrose
 ------------------ x ------------- x ------------------ = 8.5g dextrose
             1                  1L             100mL solution

likewise
    1.7L solution     10*100mL      5g dextrose
 ------------------ x ------------- x ------------------ = 85g dextrose
             1                  1L             100mL solution

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