George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil. Before adding the pasta, he adds 58 g of table salt to the water and again brings it to a boil. What is the temperature of the salty boiling water?

The idea here is that adding salt, sodium chloride, `"NaCl"`, to the pure water will increase its boiling point.

In order to determine the boiling point of the solution, you need to use the equation for boiling-point elevation, which looks like this

`color(blue)(DeltaT_b = i * K_b * b)" "`, where

`DeltaT_b` - the boiling-point elevation;
`i` - the van't Hoff factor
`K_b` - the ebullioscopic constant of the solvent;
`b` - the molality of the solution.

A quick search online will reveal that the ebullioscopic constant for water is equal to `0.512^@"C kg mol"^(-1)`

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

In order to find the value of `DeltaT_b`, which tells you the boiling point elevation of the solution, you need to know the values of the van't Hoff factor and the molality of the solution.

Now, sodium chloride is an electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations and chloride anions

`"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)`

Notice that one mole of sodium chloride produces two moles8* of ions in solution, one mole of sodium cations and one mole** of chloride anions.

This means that the van't Hoff factor, which tells you the ratio between the number of moles of solute added to the solvent and the number of moles of particles produced in solution, will be equal to `2`.

The molality of the solution tells you how many moles of solute you get per kilogram of solvent.

Use sodium chloride's molar mass to figure out how many moles are present in that sample

`58 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.9925 moles NaCl"`

This means that the molality of the solution will be

`b = "0.9925 moles"/"4.01 kg" = "0.2475 mol kg"^(-1)`

Plug in these values into the equation for boiling-point elevation and solve for `DeltaT_b`

`DeltaT_b = 2 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2475 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))`

`DeltaT_b = 0.2534^@"C"`

Now, boiling-point elevation is calculated as

`color(blue)(DeltaT_b = T_b - T_b^@)" "`, where

`T_b` - the boiling point of the solution
`T_b^@` - the boiling point of the pure solvent

Pure water boils at `100^@"C"`, which means that you have

`T_b = 100^@"C" + 0.2534^@"C" = color(green)(100.25^@"C")`

I won't round the answer off to two sig figs, despite the fact that you only have two sig figs given for the mass of sodium chloride.