A sugar exist in two isomeric form `alpha-D` & `beta-D` with specific rotation of `110^(@)` and `30^(@)` respectively. When `alpha-D` form of the suar is dissolved in water exhibits mutarotation till the equilibrium value of `90^(@)` is attained. The equilibrium constant of mutarotation of `beta-D` form [Assuming only `alpha` and `beta` form in solution] is equal to:

Answer

`00003.33`

Explanation

`alpha-Dtox%, :.90^(@)=((x)(110)+(1-x)30^(@))/`
`beta-Dto(1-x)`
`90^(@)=(110x-30x)+30^(@)`
`x=60/80=6/8`
`underset(1-x)(alpha-D)hArrunderset(x)(beta-D)`
`K_(eq)=x/(1-x)`
`:.K_(eq)^(')` for `beta-DhArralpha-D` is `1/(K_(eq))=(1-x)/x`
`:. =K_(eq)^(')=(2//8)/(6//8)=2/6=1/3=3.33`