Questions: Changes in state are called phase changes. A phase change is a change from one physical state (gas, liquid, or solid) to another. The amount of energy necessary to melt 30 kJ 1 mole of any solid is called its enthalpy of fusion. The amount of energy necessary to vaporize 1 mole of any substance is called its enthalpy of vaporization. 68 kJ The enthalpy of vaporization of liquid isopropyl alcohol is 45 kJ / mol. Calculate the energy required to vaporize 39.5 g of this compound.

Changes in state are called phase changes. A phase change is a change from one physical state (gas, liquid, or solid) to another. The amount of energy necessary to melt 30 kJ 1 mole of any solid is called its enthalpy of fusion. The amount of energy necessary to vaporize 1 mole of any substance is called its enthalpy of vaporization. 68 kJ

The enthalpy of vaporization of liquid isopropyl alcohol is 45 kJ / mol. Calculate the energy required to vaporize 39.5 g of this compound.

Transcript text: Changes in state are called phase changes. A phase change is a change from one physical state (gas, liquid, or solid) to another. The amount of energy necessary to melt $30 . \mathrm{kJ}$ 1 mole of any solid is called its enthalpy of fusion. The amount of energy necessary to vaporize 1 mole of any substance is called its enthalpy of vaporization. 68 kJ The enthalpy of vaporization of liquid isopropyl alcohol is $45 \mathrm{~kJ} / \mathrm{mol}$. Calculate the energy required to vaporize 39.5 g of this compound.

Solution

Solution Steps

Step 1: Determine the Molar Mass of Isopropyl Alcohol

To calculate the energy required to vaporize a given mass of isopropyl alcohol, we first need to determine its molar mass. The chemical formula for isopropyl alcohol is $ \text{C}_3\text{H}_8\text{O} $.

Carbon (C): $ 3 \times 12.01 \, \text{g/mol} = 36.03 \, \text{g/mol} $
Hydrogen (H): $ 8 \times 1.008 \, \text{g/mol} = 8.064 \, \text{g/mol} $
Oxygen (O): $ 1 \times 16.00 \, \text{g/mol} = 16.00 \, \text{g/mol} $

Adding these together gives the molar mass of isopropyl alcohol: \[ \text{Molar mass} = 36.03 + 8.064 + 16.00 = 60.094 \, \text{g/mol} \]

Step 2: Calculate the Number of Moles

Next, we calculate the number of moles of isopropyl alcohol in 39.5 g using its molar mass: \[ \text{Number of moles} = \frac{39.5 \, \text{g}}{60.094 \, \text{g/mol}} \approx 0.6572 \, \text{mol} \]

Step 3: Calculate the Energy Required for Vaporization

The enthalpy of vaporization of isopropyl alcohol is given as $ 45 \, \text{kJ/mol} $. Therefore, the energy required to vaporize 0.6572 moles is: \[ \text{Energy required} = 0.6572 \, \text{mol} \times 45 \, \text{kJ/mol} \approx 29.574 \, \text{kJ} \]