Question

The heat of vaporization of water is 40.66 kj/mol. how much heat is absorbed when 2.46 g of water boils at atmospheric pressure?

Answer 1

Change mass to moles. There are 18 grams in one mole of water.

2.45g / 18g = .14 moles of water

The heat absorbed is 40.66 kJ for each mol so

40.66 x .14 = 5.5 kJ 

Answer 2

Answer: The amount of heat absorbed by water is 5570.42 J

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 2.46 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{2.46g}{18g/mol}=0.137mol$

To calculate the heat absorbed by water, we use the equation:

$\Delta H_{vaporization}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = ? J

n = number of moles of water = 0.137 moles

$\Delta H_{vaporization}$ = heat of vaporization of water = 40.66 kJ/mol = 40660 J/mol    (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

$40660J/mol=\frac{q}{0.137mol}\\\\q=(40660J/mol\times 0.137mol)=5570.42J$

Hence, the amount of heat absorbed by water is 5570.42 J