# A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from 23.0°C to 18.4°C. What is the molar enthalpy of dissolution for ammonium nitrate?

##### 1 Answer

#### Explanation:

Even before doing any calculations, you can look at the change in temperature measured for the water to say that you can expect the *standard molar enthalpy of dissolution* of ammonium nitrate to be **positive**.

Since adding the ammonium nitrate salt to the water results in a **decrease in temperature**, you can conclude that the dissolution of ammonium nitrate **absorbs heat** from the surroundings

So, your strategy here will be to use the mass of the water, its specific heat, and its change in temperature to determine how much heat was absorbed by the dissolution reaction.

Your go-to equation will be

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

*final temperature* minus *initial temperature*

Plug in your values to get

#q = 60.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (18.4 - 23.0)color(red)(cancel(color(black)(""^@"C")))#

#q = -"1153.7 J"#

Now, don't be confused by the *negative sign*. If you look at things from **water's perspective**, you can say that water **gives off** heat, hence the negative sign.

Now, the standard molar enthalpy of dissolution is usually expressed in *kilojoules per mole*. Use ammonium nitrate's molar mass to determine how many moles you get in that sample

#3.88 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.04color(red)(cancel(color(black)("g")))) = "0.04848 moles NH"_4"NO"_3#

Now, if the dissolution of **one mole** of the compound will require

#1color(red)(cancel(color(black)("mole"))) * "1153.7 J"/(0.04848color(red)(cancel(color(black)("moles")))) = "23797.4 J"#

Expressed in *kilojoules per mole* and rounded to three sig figs, the answer will be

#DeltaH_"sol"^@ = color(green)(+"23.8 kJ/mol")#