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?

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 #-># you're dealing with an endothermic reaction.

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

#q# - heat absorbed/lost
#m# - the mass of the sample
#c# - the specific heat of water, equal to #4.18 "J"/("g" ""^@"C")#
#DeltaT# - the change in temperature, defined as 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 #0.04848# moles of ammonium nitrate required #"1153.7 J"# of heat, it follows that 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")#