Question

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

Answer 1

Answer:

$21.994KJ$

Explanation:

The acetone has the following molecular formula :

C3H6O

The molar mass of the acetone is

μ(C3H6O) = $58.08 \frac{g}{mole}$

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 ⇒

$x=\frac{(43.9)}{(58.08)}=0.7558mole$

In 43.9 g of C3H6O there is 0.7558 moles of C3H6O

The LHV for C3H6O is $29.1\frac{KJ}{mole}$

You can find this value in any table

To find the amount of heat :

$Q=(n).(LHV)=(0.7558mole)(29.1\frac{KJ}{mole})=21.994KJ$

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