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.

The specific rotation of alpha- glucose is +112^(@) and beta- glucose is +19^(@) and the specific rotation of the constant equilibrium mixture is +52.7^(@) . Calculate the percentage composition of anomers (alpha and beta) in the equilibrium mixture.

alpha-D -glucose hArrbeta-D glucose, equilibrium constant for this is 1.8. calculate the percentage of alpha- D glucose at equilibrium.

An organic compound P exists in two enantiomeric forms, which have specific optical rotation values [alpha] = +-100^(@) . The optical rotation of a mixture of these two enantiomers is -50^(@) . Calculate the percentage of that enantiomer which is in lower concentration in the mixture.

Glucose shows mutarotation when it dissolves in water. The specific rotation of alpha -D glucose and beta -D- glucose is +112.2^(@) and +18.7^(@) respectively. Calculate the percentage of two anomers present at equilibrium mixture with a specific rotation of +52.6^(@) .

Specific rotation for alpha -anomer of a given mono saccharide is 100^(@) and for beta -anomer is +20^(@) , specific rotation of equilibrium mixture is 68^(@) , if % of alpha anomer is x xx 10% then 'x' is