# One mole of NaCl is added to 100 mL of water in beaker A. One mole of glucose (C6H12O6) is added to 100 mL of water in beaker B. How much more will the freezing point of water be lowered in beaker A than in beaker B? Thank you!

Mar 16, 2017

The freezing point depression of $\text{beaker A}$
should be TWICE that observed for $\text{beaker B}$.

#### Explanation:

$\text{Freezing point depression}$ is a colligative property, and is proportional the $\text{number of particles in solution}$. Now glucose is not an electrolyte, and it will depress the freezing point by some factor proportional to the number of particles in solution.

On the other hand, the salt in $\text{beaker B}$ will also depress the freezing point of water by that same factor, but proportional to the number of particles in $\text{beaker A}$. But salt is a strong electrolyte, it reacts in aqueous solution to give equimolar solutions of $N {a}^{+} \left(a q\right)$, and $C {l}^{-} \left(a q\right)$.

All things being equal, $\Delta {T}_{f} \left(\text{salt solution")=2xxDeltaT_f("glucose solution}\right)$. Claro?