The standard molar heat of vaporization for water is 40.79 kJ/mol. How much energy would be required to vaporize 30.0 g of water? Answer in units of kJ.

The energy of vaporizing one mol is 40.79. 1 mol of water is 18g so the number of moles you have is 30/18. Once you know the number of moles you will need to multiply by the heat of vaporization per mole. Also water is not exactly 18g/mol so you might have to calculate it to a couple decimal places depending on what your teacher wants.

Answered by dlpeckham | 2024-06-10

The **amount **of **heat **required to **vaporize **30 g of **water **sample has been **69.98 kJ. **
**Standard **molar **heat **of **vaporization **has been given as the **amount **of heat required to **vaporize **1 **mole **of sample .
The **molar mass **of water has been 18 g. Thus, the **standard **heat of **vaporization **for water has been the **amount **of **heat **required to vaporize 18 g of water.
Computation for Heat of vaporization
The **heat **of **vaporization **of **water **has been 40.79 kJ/mol. It has been required to vaporize 18 g of sample.
The **amount **of **heat **required to **vaporize **30 g of **sample **has been:
18 g = 40.79 kJ 30 g = 18 40.79 ​ × 30 kJ 30 g = 67.98 kJ
The **amount **of **heat **required to **vaporize **30 g of **water **sample has been **69.98 kJ. **
Learn more about **heat **of vaporization , here:
https://brainly.com/question/15140455

Answered by shrutiagrawal1798 | 2024-06-15

To vaporize 30.0 g of water, approximately 68.04 kJ of energy is required. This is calculated using the standard molar heat of vaporization and the molar mass of water. The steps include calculating the number of moles in 30.0 g and then multiplying by the heat of vaporization value.
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Answered by shrutiagrawal1798 | 2024-10-01