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The specific rotation of two glucose ano...

The specific rotation of two glucose anomers are `alpha = + 110^(@)` and `beta = 19^@` and for the constant equilibrium mixtures is `+52.7^@`. Calculate the percentage compositions of the anomers in the equilibrium mixture.

Text Solution

Verified by Experts

Let a and b are mole fractions of the `alpha and beta` anomers in the equilibrium mixture.
Thus a+b =1
`110a + 19.2b = 52.7`
a+b =1
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