Question

How much heat is required to vaporize 43.9 g of acetone at its boiling point?

Answer 1

`21.994KJ`

Step-by-step explanation:

The acetone has the following molecular formula :

C3H6O

The molar mass of the acetone is

μ(C3H6O) =
`58.08 (g)/(mole)`

You can find this value in any table.

The molar mass means the mass that has 1 mole of C3H6O

Now, the heat follows this equation :

`Q=(n).(LHV)`

Where Q is the heat, n is the number of moles of the substance and LHV is the latent heat of vaporization of the substance.

Let's calculate n

If 1 mole of C3H6O has a mass of 58.08 g ⇒

x moles of C3H6O will have a mass of 43.9 g ⇒

`x=((43.9))/((58.08))=0.7558mole`

In 43.9 g of C3H6O there is 0.7558 moles of C3H6O

The LHV for C3H6O is
`29.1(KJ)/(mole)`

You can find this value in any table

To find the amount of heat :

`Q=(n).(LHV)=(0.7558mole)(29.1(KJ)/(mole))=21.994KJ`

We find out that 21.994 KJ are required to vaporize 43.9 g of acetone (C3H6O).

Answer 2

The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below

the heat of vaporization of acetone at its boiling point is 29.1 kj/mole

find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles

heat (Q) = moles x heat of vaporization

= 29.1 kj/mole x 0.757 moles = 22.03 kj