Question

the molar heat of vaporization of hydrogen fluoride (hf) is 31.18 kj/mol. how much heat is released when 100.0 g hf condenses at its boiling point? use the formula q = nδh.
+155.8 kj
-155.8 kj
-163.2 kj
+163.2 kj

Explanation:

Step1: Calculate the number of moles of HF

The molar mass of HF is $M = 1.01+19.00=20.01$ g/mol. The number of moles $n=\frac{m}{M}$, where $m = 100.0$ g. So $n=\frac{100.0}{20.01}\approx5.00$ mol.

Step2: Determine the heat released

The molar heat of vaporization $\Delta H = 31.18$ kJ/mol. When HF condenses, the process is the reverse of vaporization, so the heat change has the opposite sign. Using the formula $q=n\Delta H$, with $n = 5.00$ mol and $\Delta H=- 31.18$ kJ/mol, we get $q=5.00\times(-31.18)= - 155.9\approx - 155.8$ kJ.

Answer:

B. - 155.8 kJ