# Question c6afd

Jun 16, 2015

The boiling point of that solution is ${84}^{\circ} \text{C}$.

#### Explanation:

The equation for boiling point elevation looks like this

$\Delta {T}_{\text{b}} = i \cdot {K}_{b} \cdot b$, where

$\Delta {T}_{\text{b}}$ - the poiling point elevation;
$i$ - the van't Hoff factor;
${K}_{b}$ - the ebullioscopic constant;
$b$ - the molality of the solution.

In your case, you know that you have a solution that has a molality of 4 molal an that the ebullioscopic constant of th solvent, ${K}_{b}$, is equal to ${3.51}^{\circ} \text{C/molal}$.

Since no mention of what your solute is, you can assume that the van't Hoff factor will be equal to 1, which implies that the solute is not an electrolyte.

Plug in your values into the equation and solve for $\Delta {T}_{\text{b}}$

$\Delta {T}_{\text{b" = 1 * (3.51^@"C")/cancel("molal") * 4cancel("molal") = 14.04^@"C}}$

This means that your solution's boiling point will be ${14.04}^{\circ} \text{C}$ higher than the boling point of the pure solvent.

Therefore,

$\Delta {T}_{\text{b" = T_"b sol" - T_"b pure}}$

${T}_{\text{b sol" = 14.04 + 70.4 = 84.44^@"C}}$

I will leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the molality of the solution. Thus, the answer will be

T_"b sol" = color(green)(84^@"C")#