How many grams of salt must be added to 300g of water to increase its boiling point by 5°C?

The formula for boiling point elevation is

#color(blue)(|bar(ul(ΔT_"b" = iK_"b"m)|)#

where

• #i# is the van't Hoff factor
• #ΔT_"b"# is the boiling point elevation
• #K_"b"# is the boiling point elevation constant
• #m# is the molality of the solution

We can solve the equation for the molality of the solution.

#m = (ΔT_"b")/(iK_"b")#

In this problem,

#ΔT_"b" = "5 °C"#
#i = 2#, because 1 mol of #"NaCl"# gives 2 mol of particles in solution.
#K_"b" = "0.512 °C·kg·mol"^"-1"#

∴ #m = (5 color(red)(cancel(color(black)("°C"))))/(2 × 0.512 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "4.88 mol·kg"^"-1"#

Next, we calculate the moles of #"NaCl"#.

The formula for molality is

#color(blue)(bar(ul(|color(white)(a/a) "Molality" = "moles of solute"/"kilograms of solvent"color(white)(a/a)|)))" "#

∴ #"Moles of NaCl" = 0.300 color(red)(cancel(color(black)("kg"))) × "4.88 mol"·color(red)(cancel(color(black)("kg"^"-1"))) = "1.46 mol"#

Finally, convert moles of #"NaCl"# to grams of #"NaCl"#.

#"Mass of NaCl" = 1.46 color(red)(cancel(color(black)("mol NaCl"))) × "58.44 g NaCl"/(1 color(red)(cancel(color(black)("mol NaCl")))) = "86 g NaCl"#