Question

How much heat is absorbed when 52.3 g of H₂O(l) at 100°C and 101.3 kPa is converted to steam at 100°C? (The molar heat of vaporization of water is 40.7 kJ/mol.)

A. 1.18 x 10² kJ
B. 2.09 x 10² kJ
C. 1.11 x 10² kJ
D. 2.31 x 10¹ kJ

Answer 1

The amount of heat absorbed when 52.3 g of H₂O(l) at 100°C and 101.3 kPa is converted to steam at 100°C is approximately 118.03 kJ.

Step-by-step explanation:

To determine the amount of heat absorbed when 52.3 g of H₂O(l) at 100°C and 101.3 kPa is converted to steam at 100°C, you can use the equation:

Heat absorbed = mass of water (g) x molar heat of vaporization (kJ/mol)

First, you need to convert the mass of water to moles. The molar mass of water is 18.015 g/mol, so 52.3 g of water is equal to 52.3 g / 18.015 g/mol = 2.906 mol.

Next, multiply the moles of water by the molar heat of vaporization to find the heat absorbed:

Heat absorbed = 2.906 mol x 40.7 kJ/mol = 118.0282 kJ.