# Boiling point elevation with multiple solutes

I've encountered a problem where glucose and $$\ce{NaCl}$$ are mixed with water and $$\Delta T$$ is to be calculated.

I've calculated $$\Delta T$$ for both glucose ($$i = 1$$) and $$\ce{NaCl}$$ ($$i = 2$$) separately and added them.

Is this procedure correct or should I have treated them as one solute and used $$i = 3$$ as the Van't Hoff factor?

• You cannot use i=3 unless both solutes have the same concentration, your initial method is correct assuming you have done the calculations properly. Commented Aug 10, 2020 at 9:53
• why not ? doesn't ΔT only depend on the molality of solutes regardless what they are or their concentrations ?
– lohe
Commented Aug 10, 2020 at 10:00
• It does matter on their concentrations, you wouldn't need to care about the what the particles are, but you need to consider the number of the particles.. molality is a metric of concentration btw ;-). You would get the same answer if you were to find the $\Delta T$ of each individual ion with i=1 and add all the $\Delta T$ Commented Aug 10, 2020 at 10:10
• @The99sLearner the vant hoff factors. Commented Aug 10, 2020 at 10:46
• @The99sLearner That is not "take i=3" which is what OP later on understood. you can "take i=3" only when b(glucose) = b($\ce{NaCl}$).. Commented Aug 10, 2020 at 10:59

When there is more than $$1$$ solute in the solution, the total elevation in boiling point ($$\Delta T_\mathrm{b,total}$$) or the depression in freezing point ($$\Delta T_\mathrm{f,total}$$) is defined as the sum of each individual value for each type of solute.

What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as:

$$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{1} \label{1}$$

Now, why does this work? Let's take an example to see how it works.

We take a simple case as what you have mentioned in the question:

Find the elevation in boiling point of a solution of $$\pu{1 m}$$ glucose and $$\pu{2 m}$$ $$\ce{NaCl}$$ [$$100$$% dissociated] in water.

$$\mathrm k_\mathrm {b,water}= \pu{0.52 ^\circ C kg mol-1}$$

We can do this in two ways. One, where we take $$i = 2$$ for $$\ce{NaCl}$$. Second, we state that $$\ce{NaCl}$$ dissociates into $$\ce{Na+}$$ and $$\ce{Cl-}$$ and take the three types of solutes to be present instead of two [going from basic principles].

### Method $$1$$

Here we would have two different solute particles. Finding the value of $$\Delta T_\mathrm{b}$$ for both glucose and $$\ce{NaCl}$$ using the formula $$\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$$ we get the net value to be:

\begin{align} \Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,NaCl} \\ &= \mathrm{k_b}[1\cdot1 \,+2\cdot2] \\ &= 5\cdot\mathrm{k_b} \\ \end{align}

### Method $$2$$

Here we would have three different solute particles. Here, $$\ce{Na+}$$and $$\ce{Cl-}$$ would have a value of $$i=1$$ and their individual molality would be identical to that of $$\ce{NaCl}$$ since the salt was $$100$$% dissociated in solution

Finding the value of $$\Delta T_\mathrm{b}$$ for glucose, $$\ce{Na+}$$ and $$\ce{Cl-}$$ using the formula $$\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$$ we get the net value to be:

\begin{align} \Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,Cl^-} +\Delta T_\mathrm{b,Na^+} \\ &= \mathrm{k_b}[1\cdot1 \,+1\cdot2 + 1\cdot2] \\ &= 5\cdot\mathrm{k_b} \\ \end{align}

Therefore, both methods yield the same answer. The second method is actually the derivation for value of $$i$$ for a dissociating salt.

Now, you can apply $$i_\text{total}=i_1+i_2$$ only when the concentrations of the given solutes are the same. The proof is similar to how the second method proves the first above. In such a scenario, you can assume them to be one solute since colligative properties don't care about how many types of solute there are but rather care only about the amount of solute particles present in the solution.