Question

A 2% by weight aqueous solution of NaCl (MW= 58.45) has a boiling point of 100.31 °C. What is the van't Hoff factor for NaCl?

Answer 1

The experimental value of the van't Hoff factor is 1.7.

The formula for boiling point elevation is

`color(blue)(|bar(ul(color(white)(a/a) ΔT_b = iK_bm color(white)(a/a)|)))" "`

This can be rearranged to give

`i = (ΔT_b)/(K_bm)`

`ΔT_b = "(100.31 - 100.00) °C" = "0.31 °C"`

`K_b = "0.512 °C·kg·mol"^"-1"`

We must now calculate the molality of the solution.

Calculating the molality

Assume that we have 100 g of the solution.

Then we have 2 g of NaCl and 98 g of water.

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" = 2 color(red)(cancel(color(black)("g NaCl"))) × "1 mol NaCl"/(58.45 color(red)(cancel(color(black)("g NaCl")))) = "0.0342 mol NaCl"`

`"mass of water" = "98 g" = "0.098 kg"`

∴ `"molality" = "0.0342 mol"/"0.098 kg" = "0.349 mol/kg"`

Now we can calculate the `i` factor.

Calculating the `i` factor

`i = (ΔT_b)/(K_bm) = (0.31 color(red)(cancel(color(black)("°C"))))/(0.512 color(red)(cancel(color(black)("°C·kg·mol"^"-1"))) × 0.349 color(red)(cancel(color(black)("mol·kg"^"-1")))) = 1.7`

The van't Hoff factor for NaCl is 1.7.