Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# 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.