# What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (K_b = 0.512 C/m; molar mass of water = 18 g)?

Apr 20, 2018

$\text{101.536 °C}$

#### Explanation:

When a solute is added in a solvent the boiling of solvent is elevated.

This elevation in boiling point ($\Delta {\text{T}}_{b}$) is given by the formula

Δ"T" = i"K"_b"m"

Where

• $i =$ Van’t Hoff factor ($2$ for $\text{KBr}$)
• ${\text{K}}_{b} =$ Boiling point constant ($\text{0.512 °C/molal}$ for water)
• $\text{m =}$ Molality of solution$= \text{Moles of solute"/"Mass of solvent (in kg)}$

Here solute is glucose & solvent is water.
Boiling point of pure water is $\text{100°C}$

$\left(\text{T" - "100°C") = cancel"2" × "0.512°C/molal" × "3.00 mol"/(cancel"2.00" "kg}\right)$

$\text{T = 100°C + 1.536°C = 101.536°C}$