# What is the change in freezing point when "54 g" of glucose (use "FW" = "180 g/mol") is dissolved in "0.250 kg" of water? K_f = 1.86^@ "C/m" for water.

##### 1 Answer
May 16, 2017

$- {2.23}^{\circ} \text{C}$

Well, the fact is, we (usually) don't care what the solute is. All we needed was how many $\text{mols}$ of it we have (which we can get regardless of what its name is), whether it's a strong electrolyte or not, and how many $\text{kg}$ of water there are.

The decrease in freezing point is given by:

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - i {K}_{f} m$

where:

• ${T}_{f}$ is the freezing point of the solution.
• ${T}_{f}^{\text{*}}$ is the freezing point of the pure solvent.
• $i$ is the van't Hoff factor of the solute, i.e. the number of particles per formula unit that have gone into solution upon any dissociation.
• ${K}_{f} = {1.86}^{\circ} \text{C/m}$, or $\text{^@ "C"cdot"kg/mol}$ is the freezing point depression constant of water.
• $m$ is the molality of the solution in $\text{mol solute/kg solvent}$.

Given the mass of the glucose, we have:

54 cancel"g solute" xx "1 mol"/(180 cancel"g glucose")

$=$ $\text{0.30 mols glucose}$

And we know that $\text{250 g}$ is $\text{0.250 kg}$ of water. Water is, of course, the solvent. So:

$m = \text{0.30 mols glucose"/"0.250 kg water}$

$=$ $\text{1.20 mol/kg}$

Therefore, noting that glucose is a NONelectrolyte (i.e. its van't Hoff factor is $1$), the change in freezing point is:

$\Delta {T}_{f} = - \left(1\right) \left({1.86}^{\circ} \text{C"cdot"kg/mol")("1.20 mol/kg}\right)$

$= - {2.23}^{\circ} \text{C}$

Therefore, if we wished, we could easily find the new freezing point.

${T}_{f} - {T}_{f}^{\text{*}} = \Delta {T}_{f}$

$\implies \textcolor{b l u e}{{T}_{f}} = \Delta {T}_{f} + \cancel{{T}_{f}^{\text{*")^(0^@ "C}}}$

$= \textcolor{b l u e}{- {2.23}^{\circ} \text{C}}$

Naturally, it's the same value as the change in freezing point, since the freezing point of water is ${0}^{\circ} \text{C}$ at $\text{1 atm}$.