If #"NaCl"# is dissolved in water to raise its boiling point by #2.10^@ "C"#, what is the molarity? #K_b = 0.512^@ "C/m"#.

1 Answer
Truong-Son N.
Jun 28, 2017

I assume you mean molality.

The boiling point elevation #DeltaT_b# is defined in terms of the molality #m#, boiling point elevation constant #K_b#, and van't Hoff factor #i#:

#DeltaT_b = T_b - T_b^"*" = iK_bm#

The van't Hoff factor for #"NaCl"# is ideally #2# (the actual value is #1.9#), as 100% dissociation leads to two ions per formula unit, and thus the molality is:

#DeltaT_b = 102.10^@ "C" - 100.00^@ "C" = (2)(0.512^@ "C"cdot"kg/mol")m#

#=>m = (2.10cancel(""^@ "C"))/(2 cdot 0.512cancel(""^@ "C")cdot"kg/mol")#

#=>color(blue)(m = "2.051 mol solute/kg solvent")#

If you want the molarity, you'll have to provide the new volume of the solution.

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