Question

77.41g of isoprpanol (MW = 60.1) is used to disinfect a surface. Determine the heat absorbed, in kJ, by the surface when the entirety of the isopropanol evaporates. Take the heat of vaporization for isoprpanol to be 42.1 kJ/mol.

Answer 1

The heat absorbed by the surface when the entirety of the isopropanol evaporates is 54.28 kJ.

Step-by-step explanation:

The heat absorbed by the surface when the entirety of the isopropanol evaporates can be calculated using the formula:

Heat absorbed (kJ) = moles of isopropanol x heat of vaporization (kJ/mol)

To find the moles of isopropanol, we need to convert the given mass of isopropanol (77.41g) to moles using the molar mass. The molar mass of isopropanol is 60.1g/mol.

Moles of isopropanol = mass of isopropanol (g) / molar mass of isopropanol (g/mol)

Once we have the moles of isopropanol, we can calculate the heat absorbed by multiplying it with the heat of vaporization of isopropanol (42.1 kJ/mol).

Calculation:

Moles of isopropanol = 77.41g / 60.1g/mol = 1.289 mol

Heat absorbed = 1.289 mol x 42.1 kJ/mol = 54.28 kJ

Therefore, the heat absorbed by the surface when the entirety of the isopropanol evaporates is 54.28 kJ.