Question

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

Answer 1

You must add 86 g of salt.

Explanation:

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"`