Question

If you take Organic Chemistry Laboratory (CHM2211L) you will learn the term reflux. A reaction is considered to be at reflux when it is set at the boiling point of your solvent. a. In order to bring 74.81 mL of chloroform (density = 1.4832 g/cm3 , c = 0.96 J/g*K ) to reflux from room temperature (25 oC) it requires 1.46 kJ of energy. What will be the temperature (in oF) of the chloroform?

Answer 1

Answer : The temperature of the chloroform will be,
`101.67^oF`

Explanation :

First we have to calculate the mass of chloroform.

`\text{Mass of chloroform}=\text{Density of chloroform}* \text{Volume of chloroform}=1.4832g/ml* 74.81ml=110.958g`

conversion used :
`(1cm^3=1ml)`

Now we have to calculate the temperature of the chloroform.

Formula used :

`q=m* c* (T_(final)-T_(initial))`

where,

q = amount of heat or energy = 1.46 kJ = 1460 J (1 kJ = 1000 J)

`c` = specific heat capacity =
`0.96J/g.K`

m = mass of substance = 110.958 g

`T_(final)` = final temperature = ?

`T_(initial)` = initial temperature =
`25^oC=273+25=298K`

Now put all the given values in the above formula, we get:

`1460J=110.958g* 0.96J/g.K* (T_(final)-298)K`

`T_(final)=311.706K`

Now we have to convert the temperature from Kelvin to Fahrenheit.

The conversion used for the temperature from Kelvin to Fahrenheit is:

`^oC=(5)/(9)* (^oF-32)`

As we know that,
`K=^oC+273` or,
`K-273=^oC`

`K-273=(5)/(9)* (^oF-32)`

`K=(5)/(9)* (^oF-32)+273` ...........(1)

Now put the value of temperature of Kelvin in (1), we get:

`311.706K=(5)/(9)* (^oF-32)+273`

`T_(final)=101.67^oF`

Therefore, the temperature of the chloroform will be,
`101.67^oF`