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

Step-by-step 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}=(2.46g)/(18g/mol)=0.137mol`

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

`\Delta H_(vaporization)=(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=(q)/(0.137mol)\\\\q=(40660J/mol* 0.137mol)=5570.42J`

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