Respuesta :
Answer:
The molar enthalpy of evaporation of water at 100.0°C is 40 kJ
Explanation:
Calculation of the moles of [tex]H_2O[/tex] as:-
Mass = 0.702 g
Molar mass of [tex]H_2O[/tex] = 18 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{0.702\ g}{18\ g/mol}[/tex]
[tex]Moles= 0.039\ mol[/tex]
Given that:- [tex]\Delta H=+1.56\ kJ[/tex]
It means that when 0.039 moles of water undergoes evaporation, 1.56 kJ of energy is used
Also,
When 1 mole of water undergoes evaporation, [tex]\frac{1.56}{0.039}\ kJ[/tex] of energy is used
The molar enthalpy of evaporation of water at 100.0°C is 40 kJ
Molar enthalpy is a calculation of enthalpy change per amount of substance formed during the reaction of substances. It is expressed as kilo joules per moles.
The molar enthalpy can be calculated as:
[tex]\text {molar enthalphy} &= \dfrac{ \Delta H} {n}[/tex]
where n is the number of moles
For this moles of water can be calculated as:
[tex]\text {Moles} &= \dfrac{\text {mass}} {\text{molar mass of water}}[/tex]
- Where Mass = 0.702 g
- The molar mass of water would be= 18 g/mol
[tex]\text {Moles} &= \dfrac{\text {0.702 g}} {\text{18 g/mol}}[/tex]
Moles= 0.039 mol
- The amount of heat required is given,+1.56
- 1.56 kJ of energy is used to produce 0.039 moles of water vapor.
- One mole of water vapour undergoes evaporation at 100.0°C to produce energy:
[tex]\text {molar enthalphy} &= \dfrac{ \Delta H} {n}[/tex]
[tex]\text {molar enthalphy} &= \dfrac{ 1.56} {0.039}[/tex]
Therefore, molar enthalpy of water at 100.0°C is 40 kJ
To learn more about molar enthalphy follow the given link:
https://brainly.com/question/1399740
