Question

Methanol (CH3OH) has a heat of fusion of 3.16 kJ/mol. Which of the following is the heat of solidification that occurs when 64 grams of liquid methanol freezes?
-199 kJ
-6.32 kJ
199 kJ
6.32 kJ

Answer 1

To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:

Heat = -3.16 (64/32.06) = - 6.32 kJ

Hope this answers the question.

Answer 2

Answer : The correct option is, -6.32 KJ

Solution : Given,

Mass of methanol = 64 g

Molar mass of methanol = 32 g/mole

Heat of fusion = 3.16 KJ/mole

First we have to calculate the moles of methanol.

$\text{Moles of methanol}=\frac{\text{Mass of methanol}}{\text{Molar mass of methanol}}=\frac{64g}{32g/mole}=2mole$

Moles of methanol = 2 moles

Now we have to calculate the heat of solidification.

As, 1 mole of methanol contains heat = 3.16 KJ

So, 2 mole of methanol contains heat = $2\times 3.16KJ=6.32KJ$

The heat of solidification is, -6.32 KJ. The negative sign indicate the heat released in the system.

Therefore, the heat of solidification is, -6.32 KJ