# We dissolve 25*g mass of the following solutes in equivalent volumes of water. Which solutions will display the GREATEST freezing point depression?

Sep 20, 2016

Freezing point depression depends on the NUMBER of particles in solution; therefore, maximum depression is observed for the solute with least molar mass.

#### Explanation:

And the solute with least molar mass is clearly sodium fluoride:

$\text{Moles of sodium fluoride}$ $=$ $\frac{25 \cdot g}{41.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.595 \cdot m o l$.

$\text{Moles of magnesium sulfide}$ $=$ $\frac{25 \cdot g}{56.38 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.443 \cdot m o l$.

$\text{Moles of sodium bromide}$ $=$ $\frac{25 \cdot g}{102.89 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.243 \cdot m o l$.

$\text{Moles of silver bromide}$ $=$ $\frac{25 \cdot g}{187.77 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.133 \cdot m o l$. Note that silver bromide would not be soluble to any extent in this solution.

Clearly, when the given mass of sodium fluoride dissolves (which will cause a change in $p H$, why?), there are more solute particles in solution than with the other contenders. Thus sodium fluoride will exert the greatest colligative effect, and thus the greatest freezing point depression as compared to pure solvent.