The enthalpy for boiling water is 40.7 kJ per mole how much heat in kilojoules is required to boil 430.1 g of water

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 =

The given weight of water = 430.1 g

The molecular weight of water = 18 grams.

Moles of water =

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 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:

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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

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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