# Question e3ea3

May 25, 2017

(d) ${101.5}^{o} C$

#### Explanation:

We can automatically eliminate options (a) and (b), because one of the colligative properties of solutions is boiling-point elevation, meaning the boiling point will have increased (i.e. greater than ${100.0}^{o} C$).

The equation for calculating boiling-point elevation is

$\Delta {T}_{\text{b" = i * K_"b}} \cdot m$

where $\Delta {T}_{\text{b}}$ is the change in boiling point,
$i$ is called the van't Hoff factor, which for these purposes is essentially the number of dissolved species in one unit of solute (equal to $2$ for $\text{NaCl}$),
${K}_{\text{b}}$ is the molal boiling-point constant for water, which is given, and
$m$ is the molality of the solution.

Plugging in our known variables, we have

DeltaT_"b" = (2)(0.512( "^oC)/cancel(m))(1.5cancel(m)) = 1.5 ^oC

To find the new boiling point, simply add this to the boiling point of water, ${100}^{o} C$:

100 ^oC + 1.5 ^oC = color(red)(101.5 ^oC#

So option (d) is the correct answer.