Question

Conversion of 0.702 g of liquid water to steam at 100.0°C requires 1.56 kJ of heat. Calculate the molar enthalpy of evaporation of water at 100.0°C.

Answer 1

Answer:

The molar enthalpy of evaporation of water at 100.0°C is 40 kJ

Explanation:

Calculation of the moles of $H_2O$ as:-

Mass = 0.702 g

Molar mass of $H_2O$ = 18 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.702\ g}{18\ g/mol}$

$Moles= 0.039\ mol$

Given that:- $\Delta H=+1.56\ kJ$

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, $\frac{1.56}{0.039}\ kJ$ of energy is used

The molar enthalpy of evaporation of water at 100.0°C is 40 kJ