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FromTheMoon [43]
3 years ago
13

If you take Organic Chemistry Laboratory (CHM2211L) you will learn the term reflux. A reaction is considered to be at reflux whe

n 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?
Chemistry
1 answer:
shutvik [7]3 years ago
5 0

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}\times \text{Volume of chloroform}=1.4832g/ml\times 74.81ml=110.958g

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

Now we have to calculate the temperature of the chloroform.

Formula used :

q=m\times c\times (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\times 0.96J/g.K\times (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=\frac{5}{9}\times (^oF-32)

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

K-273=\frac{5}{9}\times (^oF-32)

K=\frac{5}{9}\times (^oF-32)+273  ...........(1)

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

311.706K=\frac{5}{9}\times (^oF-32)+273

T_{final}=101.67^oF

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

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