Answer:
972.73 KJ
Explanation:
Given data:
Enthalpy of boiling water = 40.7 KJ/mol
Heat required to boil 430.1 g water = ?
Solution:
First of all we will calculate the number of moles of water in 430.1 g.
Number of moles = mass/molar mass
Number of moles = 430.1 g/ 18 g/mol
Number of moles = 23.9 mol
Heat required:
23.9 mol × 40.7 KJ / 1mol
972.73 KJ
The enthalpy for the boiling of 430.1 grams of water has been 972.503 kJ.
The enthalpy of boiling has been the amount of energy required for the formation of 1 mole of compound.
The enthalpy of boiling water has been 40.7 kJ. It is the amount of energy required for the boiling of 1 mole of water.
The moles of given water can be calculated as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
The given weight of water = 430.1 g
The molecular weight of water = 18 grams.
Moles of water = [tex]\rm \dfrac{430.1}{18}[/tex]
Moles of water = 23.894 mol
The enthalpy of boiling 1-mole water = 40.7 kJ.
The enthalpy of boiling 23.894 mol water = 23.894 [tex]\times[/tex] 40.7 kJ
The enthalpy of boiling 23.894 mol water = 972.503 kJ.
The enthalpy for the boiling of 430.1 grams of water has been 972.503 kJ.
For more information about the enthalpy of boiling, refer to the link:
https://brainly.com/question/13756055
