The enthalpy of vaporization (ΔHvap) of water at 373 K is given as 40.7 kJ/mol. To calculate the heat (q) involved in the vaporization of 8.29 g of water, we need to know the number of moles of water present in this mass.
The molar mass of water is 18.015 g/mol, so the number of moles of water in 8.29 g can be calculated as follows:
n = m / M = 8.29 g / 18.015 g/mol = 0.460 mol
Next, we can calculate the heat (q) involved in the vaporization of 0.460 mol of water using the equation q = n ΔHvap:
q = n ΔHvap = 0.460 mol × 40.7 kJ/mol = 18.7 kJ
So, the heat involved in the vaporization of 8.29 g of water at 373 K is 18.7 kJ.