Question

Question #fdbbc

Answer 1

`"1526 g"`

Explanation:

The enthalpy change of vaporization, `DeltaH_"vap"`, tells you how much heat is needed in order for one mole of a substance to go from liquid at its boiling point to vapor at its boiling point.

This is why the enthalpy change of vaporization is positive.

Now, when a substance condenses, it goes from vapor at its boiling point to liquid at its boiling point. In this case, energy in the form of heat is being released.

The enthalpy change of condensation, `DeltaH_"cond"`, will thus carry a negative sign

`DeltaH_"cond" = - DeltaH_"vap"`

So, in order for one mole of water to condense, then enthalpy change of reaction must be equal to `DeltaH_"cond"`, i.e. `"40.79 kJ"` of heat must be released.

In your case, you know that `"3456 kJ"` of heat were released. Use the amount of heat released when one mole of water condenses at its boiling point to determine how many moles of water you have here

`3456color(red)(cancel(color(black)("kJ"))) * "1 mole water"/(40.79color(red)(cancel(color(black)("kJ")))) = "84.727 moles water"`

To determine how many grams of water would contain this many moles, use water's molar mass

`84.727 color(red)(cancel(color(black)("moles water"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole water")))) = "1526.36 g"`

Rounded to four sig figs, the answer will be

`m_"water" = color(green)(|bar(ul(color(white)(a/a)"1526 g"color(white)(a/a)|)))`