Question

Question #bdc30

Answer 1

Here's what I got.

Explanation:

Your task here is to convert the enthalpy of solution of potassium hydroxide, `"KOH"`, from kilojoules per mole, `"kJ mol"^(-)`, to kilojoules per gram, `"kJ g"^(-1)`.

To do that, you must essentially convert `1` mole of potassium hydroxide to grams by using the compound's molar mass.

`M_ ("M KOH") = "56.1056 g mol"^(-1)`

Now, write the enthalpy of solution expressed in kilojoules per mole as a conversion factor

`- "57.61 kJ mol"^(-1) = (-"57.61 kJ")/"1 mole KOH"`

Do the same for the molar mass, but keep in mind that because moles are added on the bottom of the previous conversion factor, you must have moles on top for the molar mass!

`"56.1056 g mol"^(-1) = "1 mole KOH"/"56.1056 g"`

Now multiply the two to get

`(-"56.76 kJ")/(color(red)(cancel(color(black)("1 mole KOH")))) * (color(red)(cancel(color(black)("1 mole KOH"))))/"56.1056 g" = (-"56.76 kJ")/("56.1056 g") = -"1.012 kJ g"^(-1)`

The answer is rounded to four sig figs.