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:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat absorbed = 155 kJ
n = number of moles = ? moles
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction = 40.7 kJ/mol
Putting values in above equation, we get:
[tex]40.7kJ/mol=\frac{155kJ}{n_{H_2O}}\\\\n_{H_2O}=\frac{155kJ}{40.7kJ/mol}=3.81mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of water = 18.02 g/mol
Moles of water = 3.81 moles
Putting values in above equation, we get:
[tex]3.81mol=\frac{\text{Mass of water}}{18.02g/mol}\\\\\text{Mass of water}=(3.81mol\times 18.02g/mol)=68.66g[/tex]
Hence, the mass of water that can be vaporized at its boiling point is 68.66 grams
Given:
Required heat to evaporate,
Now,
The number of moles evaporated by 155 kJ heat will bee:
= [tex]115\times \frac{1}{40.7}[/tex]
= [tex]2.8 \ mol[/tex]
hence,
The mass of water will be:
= [tex]Number \ of \ moles\times Molar \ mass \ of \ water[/tex]
= [tex]2.8\times 18[/tex]
= [tex]50.4 \ g[/tex]
Thus the above answer is correct.
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