Question

What is the expected boiling-point elevation of water for a solution that contains 150 g of sodium chloride dissolved in 1.0 kg of water?

Answer 1

`DeltaT_b = 2.6` `""^"o""C"`

Explanation:

We're asked to find the expected boiling point elevation of a solution given the amount of solute dissolved in water.

To do this, we'll use the equation

`DeltaT_b = imK_b`

where

• #DeltaT_b is the boiling point elevation; how much the boiling point increases

• `i` is the van't Hoff factor, which is essentially the number of dissolved particles per unit of solute (`2` in this case: one `"Na"^+` and one `"Cl"^-`)

• `m` is the molality of the solution.

Molality is given by

`"[molality](https://socratic.org/chemistry/solutions-and-their-behavior/molality)" = "mol solute"/"kg [solvent](https://socratic.org/chemistry/solutions-and-their-behavior/solvent)"`

We must convert the given mass of `"NaCl"` to moles using its molar mass (`58.44` `"g/mol"`):

`150cancel("g NaCl")((1color(white)(l)"mol NaCl")/(58.44cancel("g NaCl"))) = 2.57` `"mol NaCl"`

The molality is thus

`"molality" = (2.57color(white)(l)"mol")/(1.0color(white)(l)"kg") = color(red)(2.57m`

• `K_b` is the molal boiling point elevation constant for the solvent (water). The boiling point constant for water is `color(green)(0.512` `color(green)(""^"o""C/"m`

Plugging in known values, we have

`DeltaT_b = (2)(color(red)(2.57)cancel(color(red)(m)))(color(green)(0.512)(color(green)(""^"o""C"))/(cancel(color(green)(m)))) = color(blue)(2.6` `color(blue)(""^"o""C"`

which I'll round to `2` significant figures.