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

Question

Luna Chen · Accepted Answer

VERY LONG REPLY Finding the precise amount of ethanol in your 1-L sample should be your first step. You will receive 50 milliliters of ethanol for every 100 milliliters of solution because you are working with a 50% v/v solution. This indicates that you have #"%v/v" = "volume of ethanol"/"volume of solution" * 100# #"%50" = V_"ethanol"/"1000 mL" * 100 => V_"ethanol" = (50 * "1000 mL")/100 = "500 mL"# Utilizing the ethanol's density, calculate the grams contained in the 1-liter sample. #rho = m/V => m = rho * V# #m_"ethanol" = 0.79"g"/cancel("mL") * 500cancel("mL") = "395 g ethanol"# To find the number of moles you have, use the molar mass of ethanol. #395cancel("g") * "1 mole"/(46.068cancel("g")) = "8.57 moles ethanol"# Molarity is calculated by dividing the number of moles of solute by the number of liters of solution. #C = n/V# #C = "8.57 moles"/"1 L" = color(green)("8.6 M")# You must calculate the number of moles of water in the 1-L sample in order to obtain the mole fraction of ethanol. Naturally, the 500 mL of water in the 1-L bottle that you have in addition to the 500 mL of ethanol can be calculated using the molar mass and density of water. #m_"water" = 1"g"/cancel("mL") * 500cancel("mL") = "500 g water"# #500cancel("g") * "1 mole"/(18.015cancel("g")) = "27.8 moles water"# There will be a total of moles in the solution. #n_"total" = n_"ethanol" + n_"water"# #n_"total" = 8.57 + 27.8 = "36.4 moles"# The ethanol mole fraction is going to be #chi_"ethanol" = n_"ethanol"/n_"total" = (8.57cancel("moles"))/(36.4cancel("moles")) = color(green)("0.24")# Molality can be defined as the number of moles of solute in one kilogram of solvent. #b = n_"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 #"-10"^@"C"#.  To do that, you must ascertain whether the sample's alcohol content would cause the freezing point of water to drop sufficiently to permit the solution to stay liquid at that temperature. The freezing point depression equation is #DeltaT_f = K_f * b_F * i#, where #DeltaT_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 the substance you're working with isn't an electrolyte, the van't Hoff factor will be 1. The freezing point of your solution will therefore be #DeltaT_"f" =  "-1.86"^@"C"/cancel("molal") * 17cancel("molal") = "-31.6"^@"C"# #DeltaT_"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 #"-10"^@"C"#, since it will freeze at #"-32"^@"C"#.  SIDE NOTE: Although you only provided one sig fig for the sample volume, I should have used that instead of the two sig figs I used for all the answers.

Noah Aldrich · Answer

undefined Molarity: 10.2 mol/L
Mole fraction of ethanol: 0.493
Molality of ethanol: 13 mol/kg
Yes, you could still drink the whisky at -10°C.