# 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?

##### 1 Answer

#### Explanation:

For starters, you're going to have to *assume* that the density of water is equal to **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 theheatabsorbed#m# is themassof water#c# is thespecific heatof water, usually given as#"4.18 J g"^(-1)""^@"C"^(-1)# #DeltaT# is thechange in temperature, calculated as the differenec between thefinaland theinitialtemperature 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 **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 **given off** when **moles** of lithium chloride are dissolved in water, it follows that dissolving **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))))#