Question

A student is given an unknown, clear, colorless liquid at room temperature. The student measures the density, melting point, and boiling point of the liquid. What would the student conclude if he or she found out that the unknown had a density of .79 g/cm3 and a boiling point is 82.05°C? A the substance is t-Butanol B the substance is Isopropanol C the substance is Acetone D the substance is Methyl acetate

Answer 1

Opiton a) t-butanol

Step-by-step explanation:

The boiling point is a good property to identify substances because it is charateristic of each substance. The density is also characteristic of each one but its dependence of the temperature could lead to wrong conclusions if you don´t measure right the room temperature.

Being that said, once calculated the boiling point, in tables you will find the boiling point of many substances. By looking the value you obtained you can determine which substance it is.

So:

`Tb_(t-butanol)=82^oC`

`Tb_(isopropanol)=82.5^oC`

`Tb_(acetone)=56^oC`

`Tb_(methyl acetate)=57.1^oC`

As can be seen, the unknown substance is t-butanol