You have 1 L of 100 proof (50% V/V) Scotch whisky (ethanol, C_2H_5OH). Calculate the molarity, the mole fraction, the molality of the ethanol and if the temperature drops to -10C could you still drink the whisky? Density of ethanol =0.79g/ml kf=-1.86C/m

May 6, 2015

The first thing you need to do is determine exactly how much ethanol your 1-L sample contains.

Since you're dealing with a 50% v/v solution, you'll get 50 mL of ethanol for every 100 mL of solution, which means that you have

$\text{%v/v" = "volume of ethanol"/"volume of solution} \cdot 100$

$\text{%50" = V_"ethanol"/"1000 mL" * 100 => V_"ethanol" = (50 * "1000 mL")/100 = "500 mL}$

Use the density of ethanol to determine how many grams you have in the 1-L sample

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

${m}_{\text{ethanol" = 0.79"g"/cancel("mL") * 500cancel("mL") = "395 g ethanol}}$

Use ethanol's molar mass to determine how many moles you have

395cancel("g") * "1 mole"/(46.068cancel("g")) = "8.57 moles ethanol"

Since molarity is defined as moles of solute divided by liters of solution, you'll get

$C = \frac{n}{V}$

C = "8.57 moles"/"1 L" = color(green)("8.6 M")

To get the mole fraction of ethanol, you need to determine how many moles of water you have in the 1-L sample.

Since you have 500 mL of ethanol in the 1-L bottle, you'll of course also have 500 mL of water. Use water's density and its molar mass to determine how many moles you have

${m}_{\text{water" = 1"g"/cancel("mL") * 500cancel("mL") = "500 g water}}$

500cancel("g") * "1 mole"/(18.015cancel("g")) = "27.8 moles water"

The total number of moles present in the solution will be

${n}_{\text{total" = n_"ethanol" + n_"water}}$

${n}_{\text{total" = 8.57 + 27.8 = "36.4 moles}}$

The mole fraction of ethanol will be

chi_"ethanol" = n_"ethanol"/n_"total" = (8.57cancel("moles"))/(36.4cancel("moles")) = color(green)("0.24")

Molality is defined as moles of solute per kilogram of solvent, so you'll get

$b = {n}_{\text{ethanol"/"kg of water}}$

b = "8.57 moles"/(500 * 10^(-3)"kg") = color(green)("17 molal")

Now you have to determine whether or not you can still drink the whiskey if the temperature of the sample is dropped to $\text{-10"^@"C}$.

To do that, you have to determine if the amount of alcohol present in the sample would lower the freezing point of water enough to allow for the solution to remain liquid at that temperature.

$\Delta {T}_{f} = {K}_{f} \cdot {b}_{F} \cdot i$, where

$\Delta {T}_{f}$ - the freezing-point depression, defined as the freezing temperature of the pure solvent minus the freezing point of the solution;
${K}_{f}$ - the cryoscopic constant, which depends solely on the solvent;
${b}_{F}$ - the molality of the solution;
$i$ - the van't Hoff factor, which takes into account the number of particles produced by a compound when dissolved in solution.

Since you're dealing with a non-electrolyte, the van't Hoff factor will be equal to 1.

So, your solution will have a freezing point of

$\Delta {T}_{\text{f" = "-1.86"^@"C"/cancel("molal") * 17cancel("molal") = "-31.6"^@"C}}$

$\Delta {T}_{\text{f" = T_"f water" - T_"f solution}}$

T_"f solution" = DeltaT_"f" - DeltaT_"f water" = "-31.6"^@"C" - 0^@"C" = color(green)("-32"^@"C")

Therefore, you can still drink the whiskey at $\text{-10"^@"C}$, since it will freeze at $\text{-32"^@"C}$.

SIDE NOTE I've given all the answers with two sig figs, but I should have used 1 sig fig, since that is all you gave for the volume of the sample.