Question

Question #acab0

Answer 1

`-"13.4 kJ mol"^(-1)`

Explanation:

The problem wants you to find the enthalpy change of solution, `DeltaH_"sol"`, for aluminium sulfate, `"Al"_2("SO"_4)_3`, by using the change in the temperature of the 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 `DeltaH_"sol"` 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

`q` - the amount of heat gained
`m` - the mass of the sample
`c` - the specific heat of water
`DeltaT` - the 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 `"0.9975 g mL"^(-1)` at that temperature interval, which means that the sample will have a mass of

`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 `"25.0 g"` of aluminium sulfate are dissolved in that sample of water. To get the heat 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 `"975.7 J"` are being given off when `0.07307` moles of aluminium sulfate are dissolved in water, it follows that dissolving 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))`