A bottle of Bourbon Whiskey contains 250.ml of liquid and 42% alcohol (C2H5OH). assuming density = 1.05g/ml, will the whiskey freeze if put into a -20.0 degree celsius freezer?

Question

can someone please explain.

Answer 1

It looks like we're looking at a freezing point depression equation. For this, ∆T =K_f*m*i

m is our molality, which is the moles of solute per kg of solvent. To find the number of moles of alcohol, we multiply 250mL by 0.42 to obtain 105mL.

We can then use dimensional analysis and the molar mass of ethanol to find the number of moles of ethanol (the alcohol).

105mL * (1.05g / mL) * (1 mol / 46.07g) = 2.4 moles ethanol

To obtain kg of solvent (assuming that it's water):

250mL total solution - 105 mL of ethanol = 145 mL solvent

145mL solvent * (1.0 g / mL ) * (1kg / 1000g) = 0.15 kg solvent

m = 2.4 moles alcohol / 0.15 kg solvent = 16 molality

i = 1, since ethanol has a van't hoff factor of 1.

If we are assuming the liquid is water, we can use the K_f of water, which is 1.853 K*kg*mol^-1

∆T = T_{F(pure solvent)}-T_F(solution) . Pure solvent freezing point of water is 0C (273K), so now we just put the numbers into our equation and solve.

273K - T_solution = (1.853 K*kg*mol^-1)(1)(16mol*kg^-1)

273K - T_solution = 29.648K

T_solution = 243.35K = -29.648C

Therefore, it will not freeze since the solution will freeze at -29.6C

Note: I have made some assumptions. I assumed that the % alcohol was volume/volume % and that the liquid is water.