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

Answer:

#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.