In an experiment at constant pressure, 4.24g of lithium chloride is dissolved in 100. mL of water at initial temperature of 16.3 degrees C. The final temperature of the solution is 25.1 degrees C. What is the enthalpy of solution in kJ/mol?

For starters, you're going to have to assume that the density of water is equal to #"1.0 g mL"^(-1)# in order to be able to find the mass of water that absorbed the heat given off when your sample of lithium chloride dissolved.

#100. color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "100. g"#

Now, you can determine the heat absorbed by the water by using the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

• #q# is the heat absorbed
• #m# is the mass of water
• #c# is the specific heat of water, usually given as #"4.18 J g"^(-1)""^@"C"^(-1)#
• #DeltaT# is the change in temperature, calculated as the differenec between the final and the initial temperature of the water

In your case, you have

#q = 100. color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (25.1 - 16.3)color(red)(cancel(color(black)(""^@"C")))#

#q = "3678.4 J"#

Now, the problem wants you to find the molar enthalpy of solution, which basically means that you must find the change in enthalpy that occurs when #1# mole of lithium chloride dissolves.

Use the molar mass of lithium chloride to convert the mass of the sample to moles

#4.24 color(red)(cancel(color(black)("g"))) * "1 mole LiCl"/(42.394color(red)(cancel(color(black)("g")))) ~~ "0.100 moles LiCl"#

So, if #"3678.4 J"# of heat are being given off when #0.100# moles of lithium chloride are dissolved in water, it follows that dissolving #1# mole will give off

#1 color(red)(cancel(color(black)("mole LiCl"))) * "3678.4 J"/(0.100color(red)(cancel(color(black)("moles LiCl")))) = "36784 J"#

Finally, since this represent heat given off, the molar enthalpy of solution will be negative. Expressed in kilojoules per mole and rounded to three sig figs, the answer will be

#color(darkgreen)(ul(color(black)(DeltaH_"sol LiCl" = -"36.8 kJ mol"^(-1))))#