# Question #acab0

##### 1 Answer

#### Answer:

#### Explanation:

The problem wants you to find the *enthalpy change of solution*,

Now, the *enthalpy change of solution* represents the change in enthalpy when **one mole** of a substance dissolves in water to form a solution of infinite dilution.

An important thing to notice here is that the temperature of the solution **increases** upon dissolving the salt, which means that heat is being **given off** when aluminium sulfate dissolves in water.

Since this is characteristic of an **exothermic process**, you know right from the start that **must** carry a **negative sign**.

So, the idea here is that the heat *given off* by the dissolution of the salt will be **equal** to the heat *absorbed* by the solution.

#color(blue)(-q_"salt" = q_"water")#

You can assume that the **specific heat** of the solution will be equal to that of water

#c = 4.18"J"/("g" ""^@"C")#

The heat absorbed by the water can be calculated using the equation

#color(blue)(q = m * c * DeltaT)" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

Since the sample provides the *volume* of the water, you're going to have to use its **density** to find the mass. I will take the density to be approximately

#60.0 color(red)(cancel(color(black)("mL"))) * "0.9975 g"/(1color(red)(cancel(color(black)("mL")))) = "59.85 g"#

The water will thus absorb

#q_"water" = 59.85color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25.3 - 21.4)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = "975.7 J"#

This means that the heat **given off** by the dissolution of the salt was

#q_"salt" = -"975.7 J"#

Remember, the minus sign is used to show *heat lost*.

Now, this much heat is being given off when **one mole** dissolves, convert this to *moles* by using the compound's molar mass

#25.0color(red)(cancel(color(black)("g"))) * ("1 mole Al"_2("SO"_4)_3)/(342.15color(red)(cancel(color(black)("g")))) = "0.07307 moles Al"_2("SO"_4)_3#

So, if **given off** when **one mole** will give off

#1color(red)(cancel(color(black)("mole"))) * "975.7 J"/(0.07307color(red)(cancel(color(black)("moles")))) = "13353 J"#

Rounded to three **sig figs** and expressed in *kilojoules*, the *enthalpy change of solution* will thus be

#DeltaH_"sol" = color(green)(-"13.4 kJ mol"^(-1))#