What is the new boiling point for an aqueous "0.743 molal" solution of "KCl"? K_b = 0.512^@ "C/m"
1 Answer
Aug 9, 2017
As you should have seen in your book,
DeltaT_b -= T_b - T_b^"*" = iK_bm ,where:
DeltaT_b is the change in boiling point in""^@ "C" , from that of the pure solvent,T_b^"*" , to that of the solution,T_b .i is the van't Hoff factor, i.e. the effective number of solute particles in solution.K_b = 0.512^@ "C/m" is the boiling point elevation constant of water.m is the molality of the solution..."mol solute/kg solvent" . Is the solute volatile or nonvolatile?
Assuming
"KCl"(aq) -> "K"^(+)(aq) + "Cl"^(-)(aq)
and
color(blue)(T_b) = T_b^"*" + iK_bm
= 100^@ "C" + 2 cdot 0.512^@ "C/m" cdot "0.743 m"
= color(blue)ul(100.761^@ "C")
What was the change in boiling point?