A mixture of 1-butanol and chlorobenzene is heated and the vapours at 115.3 degrees celsius condensed and collected. What is the percent weight composition of this distillate if a careful fractional distillation was performed?
Wouldn't it depend on the ratios of vapor pressures of each at that temp?
To determine the percent weight composition of the distillate, we need to know the vapor pressure of each component at the given temperature and the mol fraction of each component in the liquid mixture.
First, let's consider the vapor pressure of 1-butanol and chlorobenzene at 115.3 degrees Celsius. You can find these values in a reference handbook, chemical database, or literature.
Next, fractional distillation separates the components based on their boiling points. Since 1-butanol has a lower boiling point than chlorobenzene, it will preferentially evaporate first.
During fractional distillation, the vapor composition changes as the distillation progresses. The composition will gradually become richer in the lower boiling component (1-butanol) and lighter in the higher boiling component (chlorobenzene).
To determine the percent weight composition, you need to know the mol fraction of each component in the liquid mixture before distillation. The mol fraction can be calculated using mole ratios or measured experimentally using a technique such as gas chromatography.
Once you have the mol fractions of 1-butanol and chlorobenzene in the liquid mixture, you can calculate the weight percent composition of the distillate using the following formula:
Weight percent of 1-butanol = (Mol fraction of 1-butanol) * 100
Weight percent of chlorobenzene = (Mol fraction of chlorobenzene) * 100
Note: It's important to perform careful measurements and calculations to ensure accuracy in determining the weight percent composition.