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?

1 Answer
Jul 9, 2017

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" = "mol solute"/"kg 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.