D-Glucose Equilibrium in Aq Soln

[pancakes]

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.

[adianadiadi]

Will you post the exact problem. Your question is not clear. How do you figured out there are 100 g of glucose?