Question

Calculate the mass of water (in g) that can be vaporized at its boiling point with 155 kJ of heat. • ΔHvap = 40.7 kJ/mol (at 100 °C) • 18.02 g H2O = 1 mol H2O

Answer 1

Answer: The mass of water that can be vaporized at its boiling point is 68.66 grams

Explanation:

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = 155 kJ

n = number of moles = ? moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = 40.7 kJ/mol

Putting values in above equation, we get:

$40.7kJ/mol=\frac{155kJ}{n_{H_2O}}\\\\n_{H_2O}=\frac{155kJ}{40.7kJ/mol}=3.81mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of water = 18.02 g/mol

Moles of water = 3.81 moles

Putting values in above equation, we get:

$3.81mol=\frac{\text{Mass of water}}{18.02g/mol}\\\\\text{Mass of water}=(3.81mol\times 18.02g/mol)=68.66g$

Hence, the mass of water that can be vaporized at its boiling point is 68.66 grams