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
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
where
-
#DeltaT_b is the boiling point elevation; how much the boiling point increases
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#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
which I'll round to