Question

The heat of vaporization of water is 40.66 KJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?

Answer 1

Hint: To calculate the heat absorbed, first we need to find out the number of moles of water by dividing the mass of the water with the molecular weight of the water. The molar heat of vaporization is the energy required to vaporize one mole of liquid.

Complete step by step answer:
It is given that the heat of vaporization of water is 40.66 KJ/mol.
The mass of water is 2.87 g.
The molar heat of vaporization is defined as the energy required to vaporize one mole of liquid. The unit of molar heat of vaporization is kilojoule per mole KJ/mol. The molar heat of vaporization is denoted by $\Delta {H_{vap}}$. The molar heat of vaporization is sometimes referred to as molar enthalpy of vaporization.
Here, the heat of vaporization of water is 40.66 KJ/mol implies that we need 40.66 KJ/mol of heat to boil 1 mole of water at 100 degree Celsius, normal boiling point temperature.
First we need to calculate the number of moles from the given mass. The molecular weight of water is 18.015 g/mol.
The formula is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molecular weight
To calculate the number of moles, substitute the values in the above equation.
$ \Rightarrow n = \dfrac{{2.87g}}{{18.015g/mol}}$
$ \Rightarrow n = 0.1593mol$
Now use the value of molar heat of vaporization as the conversion factor to find out how much heat is needed to boil 0.1593 mol of water.
$ \Rightarrow 0.1593mols\;of\;{H_2}O \times \dfrac{{40.66KJ}}{{1\;mol\;of\;{H_2}O}}$
$ \Rightarrow 6.477KJ \approx 6.48KJ$
Therefore, 6.48 KJ heat is absorbed when 2.87 g of water boils at atmospheric pressure.

Note:
Make sure to convert the calculated value to 3 significant figures which is the number of significant figures of the mass of the water.