73.7k views
4 votes
The heat of vaporization for water is 40. 7 kJ/mol. A given amount of water requires 63. 8 kJ of energy to boil. The mass of the sample of water is -

13. 2 g

18. 0 g

22. 28

28. 2g

62. 38

1 Answer

3 votes

answer is B value of approximation

solving

The heat of vaporization for water is 40.7 kJ/mol, which means that it takes 40.7 kJ of energy to convert 1 mole of water from a liquid to a gas phase.

The energy required to boil the water is given as 63.8 kJ.

We can set up a proportion using these values:

40.7 kJ/1 mol = 63.8 kJ/x mol

To solve for x (the number of moles of water), we can cross-multiply and solve the equation:

40.7 kJ * x mol = 63.8 kJ * 1 mol

x mol = (63.8 kJ * 1 mol) / 40.7 kJ

x mol ≈ 1.57 mol

Since we know the molar mass of water is approximately 18.015 g/mol, we can convert moles to grams:

1.57 mol * 18.015 g/mol ≈ 28.29 g

Therefore, the mass of the sample of water is approximately 28.29 grams.

User Ricbermo
by
8.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.