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Topic: Lab Question About D-Glucose Equilibrium in Aq Soln  (Read 2616 times)

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Offline pancakes

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Lab Question About D-Glucose Equilibrium in Aq Soln
« on: August 03, 2010, 11:08:07 PM »
Three solid forms of glucose are anhydrous alpha, alpha hydrate, and anhydrous beta. Assume at equilibrium at 23, the solution is 50 wt% total glucose with beta:alpha = 5:3
Note that alpha hydrate is 10/11 glucose, e.g., 11 g hydrate corresponds to 10 g anhydrous.
Exactly 1100 g alpha hydrate is mixed with water. @ equilibrium, there is 160g aqueous solution plus solids. What is the mass of each solid phase?

The answer is supposed to be 22 g of alpha glucose hydrate. But how do you get that? I figured out that the whole system has 100g glucose on anhydrous basis and 30g of alpha and 80g of beta (80g of total glucose) in solution. But where does the 22g come from? I am so confused.

Offline Dan

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Re: Lab Question About D-Glucose Equilibrium in Aq Soln
« Reply #1 on: August 08, 2010, 11:46:41 AM »
Can you post the question exactly as it is written. It is confusing as you've written it as it seems to contain irrelevant information and the numbers look as though they don't add up.

Exactly 1100 g alpha hydrate is mixed with water.

1100 g or 110 g?

Quote
I figured out that the whole system has 100g glucose on anhydrous basis and 30g of alpha and 80g of beta (80g of total glucose) in solution.

I agree with the green bit, but the red bit makes no sense... 30+80 = 110 not 80...

If the total amount of glucose added is 100 g (based on anhydrous) and the solution has mass 160 g you can now calculate:

1. The mass of glucose in solution
2. The mass of glucose (based on anhydrous) not in solution

Now you need to know which form of glucose precipitates from water in order to calculate the mass of solid. I am surprised that this is not given in the question, so I assume you are supposed to know that it is the alpha hydrate. As a side point, the anomeric ratio of aqueous glucose is completely irrelevant as far as I can see.
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