ANS:A - The mixture exhibits an azeotrope at 101.3 kPa.

The mixture exhibits an azeotrope at 101.3 kPa: An azeotrope is a mixture of two or more components in such proportions that its composition cannot be altered by simple distillation. In this case, the mixture does not exhibit an azeotrope at 101.3 kPa because azeotropes typically occur at specific compositions and conditions where the vapor and liquid phases have the same composition. The given mixture consists of three different isomers of xylene (o-xylene, p-xylene, and m-xylene) in different proportions, which does not meet the criteria for forming an azeotrope.

1. The composition of the mixture, in percent by volume is: o-xylene 30, p-xylene 60, and m-xylene 10: This statement is not necessarily true. Percentages by weight (w/w) were provided for the mixture, but percentages by volume are not given. Without additional information about the densities of the individual components, we cannot determine the composition by volume.
2. The composition of the mixture in mole percent is: o-xylene 30, p-xylene 60, and m-xylene 10: This statement is true. Given the composition in weight percent (w/w), we can convert these percentages into mole fractions to find the composition in mole percent. The mole percent composition will be the same as the weight percent composition because the molecular weights of o-xylene, p-xylene, and m-xylene are similar.
3. The mixture contains optical isomers: This statement is false. Optical isomers, also known as enantiomers, occur when a molecule has a chiral center, resulting in non-superimposable mirror images. Xylene isomers (o-xylene, p-xylene, and m-xylene) do not have chiral centers, so they do not exhibit optical isomerism. They are structural isomers but not optical isomers.