1.

The specific rotation for the alpha-anomer of a carbohydrate is +29^(@) and that of beta-anomer is -17^(@). Due to mutarotation, the equilibrium specific rotation is +14^(@). The percentage of alpha-anomer is

Answer»

`67.4%`
`32.6%`
`70.67%`
`29.33%`

Solution :`67.4 %`
Let the MOLES of `alpha` -anomer at EQUILIBRIUM = a
The number of moles of `BETA`-anomer `=(1-a)`
`therefore""29a-17(1-a)=14`
`therefore""a=0.6739`


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