The following information is given for cadmium at 1atm:

Question

boiling point = 765 °C	Hvap(765 °C) = 100 kJ/mol
melting point = 321 °C	Hfus(321 °C) = 6.11 kJ/mol
specific heat solid= 0.230 J/g°C
specific heat liquid = 0.264 J/g°C

 kJ are required to melt a 43.1 g sample of solid cadmium, Cd, at its normal melting point.

Answer 1

Atomic mass Cd = 112.4 g/mol

moles Cd present = 43.1 g x 1 mol/112.4 g = 0.3835 moles

Since the problems states that the Cd is already at its normal melting point, the energy needed to melt the Cd will simply be the ∆H_fusion x moles...

0.3835 moles x 6.11 kJ/mole = 2.34 kJ (to 3 significant figures)

Answer 2

Like most chemistry questions, this one takes knowing a general definition (in this case, the molar heat of fusion) and applying some information specific to the situation described in the problem.

The molar heat of fusion of a substance is the amount of heat energy it will take to melt a mole of that substance, provided the sample is already at its melting point. (It's also the amount of heat energy released when a mole of that substances freezes, but that's not important here.)

Since, according to the problem, the Cd sample is already at its melting point, you would simply pay an amount of energy proportionate to the amount of the substance, and it ends up just being a conversion factor problem:

mass of Cd sample x reciprocal of the molar mass of Cd x molar heat of fusion.

You can handle that I'm sure.

You shouldn't need to worry about the specific heats, since those what it costs in heat energy to raise the temperature of a gram of a substance by a single degree-- and you aren't raising the temperature of anything.

This is where really knowing the definitions helps you avoid confusion.

The key insight is that, at any constant-pressure transition of state, all the energy you're administering to the sample is going to go into breaking down the intermolecular forces between the solid particles of Cd, such that essentially none of it is available to raise their average kinetic energy aka temperature.

Cheers!