How much heat must be removed to freeze a tray of ice cubes if the water has a mass of 40.0 g? (the molar heat of fusion of water is 6.01 kj/mol.)

We sill use the formula:

Q = nl, where Q is the heat, n is the number of moles and l is the latent molar heat of fusion of water.

moles = mass / Mr

n = 40 / 18

n = 2.22 moles

Q = 2.22 x 6.01

Q = 13.34 kJ of heat must be removed

We will use the formula Q = nl, where Q is the heat, n is the number of moles and l is the latent molar heat of fusion of water.

The formula to calculate moles is moles = mass / Mr, so n = 40 / 18 which comes out to be 2.22 moles.

Now we can calculate the heat, Q = 2.22 x 6.01. So 13.34 kJ of heat must be removed in order to freeze a tray of ice of 40 g