# A sample of substance X that has a mass of 326.0 g releases 4325.8 cal when it freezes at its freezing point. If substance X has a molar mass of 58.45 g/mol, what is the molar heat of fusion for substance X?

Jun 14, 2017

$\Delta {H}_{\text{fus" = "775.6 cal mol}}^{- 1}$

#### Explanation:

For a given substance, the molar enthalpy of fusion, $\Delta {H}_{\text{fus}}$, tells you the enthalpy change that accompanies a solid $\to$ liquid phase change of $1$ mole of said substance.

This implies that the molar enthalpy of fusion will almost always have a positive value, since heat is usually needed in order for a substance to go from solid at its melting point to liquid at its melting point.

Now, your goal here is to figure out the enthalpy change that occurs when $1$ mole of substance $\text{X}$ goes from liquid to solid at its freezing point.

You already know that the substance gives off energy when it freezes, so right from the start, you know that this enthalpy change will carry a negative sign.

DeltaH_ ("liquid " -> " solid") = -"(a value) mol "^(-1)

The minus sign symbolizes heat given off and ${\text{mol}}^{- 1}$ symbolizes that this enthalpy change occurs per mole of substance

So start by making a note that you must have

overbrace(DeltaH_"fus")^(color(blue)("must be positive")) = - overbrace(DeltaH_ ("liquid " -> " solid"))^(color(purple)("will be negative"))

Next, use the molar mass of substance $\text{X}$ to convert the mass to moles.

326.0 color(red)(cancel(color(black)("g"))) * "1 mole X"/(58.45color(red)(cancel(color(black)("g")))) = "5.5774 moles X"

So, you know that when $5.5774$ moles of $\text{X}$ freeze, $\text{4325.8 cal}$ of heat are being given off. Use this information to find the heat given off when $1$ mole of $\text{X}$ freezes

1 color(red)(cancel(color(black)("mole X"))) * "4325.8 cal"/(5.5774color(red)(cancel(color(black)("mole X")))) = "775.6 cal"

You can thus say that the enthalpy change that occurs when $1$ mole of $\text{X}$ freezes is equal to

DeltaH_ ("liquid " -> " solid") = - "775.6 cal mol"^(-1)

Therefore, the molar enthalpy of fusion will be equal to

DeltaH_"fus" = - (-"775.6 cal mol"^(-1))

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\Delta {H}_{\text{fus" = +"775.6 cal mol}}^{- 1}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the mass of the sample.