What is the boiling point of a 1.86 m aqueous solution of MgCl2, an ionic compound?

The boiling point is 102.86 °C.

We use the boiling point elevation expression

##ΔT_b = iK_bm##

where

• ##ΔT_b## = change in boiling point;
• ##i## is the van’t Hoff factor;
• ##K_b## is the molal boiling point elevation constant for the ;
• ##m## is the of the .

Step 1. Calculate the ##i## value.

##"MgCl"_2("s") → "Mg"^(2+)("aq") + "2C"l⁻("aq")##

Since 1 mol of ##"MgCl"_2## forms 3 mol of ions, ##i = 3##.

Step 2. Calculate ##ΔT_b##.

For water, ##K_b = "0.512 °C·kg·mol"^-1##

##ΔT_b = iK_bm = "3 × 0.512 °C·kg·mol"^-1 × "1.86 mol·kg"^-1 = "2.86 °C"##

Step 3.

##T_b = T_b° + ΔT_b = "(100.00 + 2.86) °C" = "102.86 °C"##

Here's a video on calculating a boiling point.