Question

What is the boiling point of a 1.5 m soloution of KCI

Answer 1

We could use the following equation for boiling point elevation:

$\Delta T=i\cdot m\cdot K_b$

Where:

$\begin{gathered} \Delta T=Change\text{ in temperature} \\ i=\text{ Van Hoff factor} \\ m=molality \\ K_b=\frac{0.51C}{m} \end{gathered}$

Now, remember that:

Since there are two elements in the compound dissociating, the Van Hoff's factor will be:

$i\to(K^+,Cl^-)=2$

We just have to replace the given values and solve for the change in temperature:

$\begin{gathered} \Delta T=(2)(1.5m)(\frac{0.51C}{m}) \\ \Delta T=1.53C \end{gathered}$

The change in temperature equals 1.53°C. As you know, the initial boiling point was 100°C, so, if temperature changed 1.53°C, the new boiling point of the solution is 101.53°C.