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
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#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" = "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 iscolor(green)(0.512 color(green)(""^"o""C/"m
Plugging in known values, we have
which I'll round to