Question

How much heat is released when 12.2 grams of methanol freezes at -268 ◦C?

A) 10.5 J
B) 25.6 J
C) 50.2 J
D) 75.8 J

Answer 1

Approximately 2.3 kJ of heat is released when 12.2 grams of methanol freezes at -268°C.

Step-by-step explanation:

To determine the amount of heat released when 12.2 grams of methanol freezes at -268℃, we need to use the specific heat of fusion of methanol. The specific heat of fusion for methanol is 6.01 kJ/mol. First, we convert the grams of methanol to moles. The molar mass of methanol is approximately 32 g/mol. So, we divide 12.2 g by 32 g/mol to get approximately 0.381 mol. Next, we multiply the moles of methanol by the molar heat of fusion. 0.381 mol multiplied by 6.01 kJ/mol gives us approximately 2.3 kJ of heat released.