# What is the expected boiling-point elevation of water for a solution that contains 150 g of sodium chloride dissolved in 1.0 kg of water?

Jul 9, 2017

$\Delta {T}_{b} = 2.6$ $\text{^"o""C}$

#### Explanation:

We're asked to find the expected boiling point elevation of a solution given the amount of solute dissolved in water.

To do this, we'll use the equation

$\Delta {T}_{b} = i m {K}_{b}$

where

• DeltaT_b is the boiling point elevation; how much the boiling point increases

• $i$ is the van't Hoff factor, which is essentially the number of dissolved particles per unit of solute ($2$ in this case: one ${\text{Na}}^{+}$ and one ${\text{Cl}}^{-}$)

• $m$ is the molality of the solution.

Molality is given by

$\text{[molality](https://socratic.org/chemistry/solutions-and-their-behavior/molality)" = "mol solute"/"kg [solvent](https://socratic.org/chemistry/solutions-and-their-behavior/solvent)}$

We must convert the given mass of $\text{NaCl}$ to moles using its molar mass ($58.44$ $\text{g/mol}$):

150cancel("g NaCl")((1color(white)(l)"mol NaCl")/(58.44cancel("g NaCl"))) = 2.57 $\text{mol NaCl}$

The molality is thus

"molality" = (2.57color(white)(l)"mol")/(1.0color(white)(l)"kg") = color(red)(2.57m

• ${K}_{b}$ is the molal boiling point elevation constant for the solvent (water). The boiling point constant for water is color(green)(0.512 color(green)(""^"o""C/"m

Plugging in known values, we have

DeltaT_b = (2)(color(red)(2.57)cancel(color(red)(m)))(color(green)(0.512)(color(green)(""^"o""C"))/(cancel(color(green)(m)))) = color(blue)(2.6 color(blue)(""^"o""C"#

which I'll round to $2$ significant figures.