cheistry help?

2016-05-17 5:55 am

The molar heat of vaporization of water is 40.7
kJ/
mol
. How much heat must be absorbed to convert 50.0 grams of liquid water at 100oC to steam at 100oC?
A) 1.46 X 10 kJ
B) 1.13 X 102kJ
C) 2.04 X 103kJ
D) 3.66 X 104kJ

回答 (1)

胡雪8°

2016-05-17 11:01 am

✔ 最佳答案

Molar mass of water (H₂O) = (1.0×2 + 16.0) g/mol = 18 g/mol

"The molar heat of vaporization of water is 40.7 kJ/mol" means
When 1 mol of liquid water is converted to steam at 100°C, 40.7 kJ of heat is needed.
OR: When 18.0 g of liquid water is converted to steam at 100°C, 40.7 kJ of heat is needed.
OR: When 1.00 g of liquid water is converted to steam at 100°C, (40.7/18.0) kJ of heat is needed.

When 50.0 g of liquid water is converted to steam at 100°C, heat needed
= (40.7/18.0) × 50.0 kJ
= 113 kJ
= 1.13 × 10² kJ

...... The answer is: C) 1.13 × 10² kJ

👍 1 👎 0