When 18 g of ethylene glycol C2H6O2 is dissolved in 150 g of pure water, the freezing point of the solution is ___ (degrees)C. ( The freezing point depression constant for water is 1.86(dgrees)C kg/mol.
I know the formulas I just need to see it done step by step please.
delta T = Kf*m
where m = molality = # mols/kg solvent. and
# mols = g/molar mass.
Excuse me. Initially I read only part of your question. If you know the formulae, then what is your problem. It's just a plug and chug.
To find the freezing point of the solution, we need to calculate the freezing point depression caused by the addition of ethylene glycol. The formula to calculate freezing point depression is:
∆Tf = Kf * m
where:
∆Tf = freezing point depression
Kf = freezing point depression constant
m = molality of the solution
First, let's determine the molality of the solution by calculating the moles of ethylene glycol and the mass of water present.
Step 1: Calculate the moles of ethylene glycol (C2H6O2):
To calculate moles (n), divide the given mass (m) by the molar mass (M) of ethylene glycol.
Given mass of ethylene glycol (m) = 18 g
Molar mass of ethylene glycol (M) = 2 * atomic mass of C + 6 * atomic mass of H + 2 * atomic mass of O
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 62.07 g/mol
n = m / M
n = 18 g / 62.07 g/mol
n ≈ 0.2904 mol
Step 2: Calculate the mass of water (solvent):
Given mass of water = 150 g
Step 3: Calculate the molality (m):
The molality (m) is defined as the amount of moles of solute per kilogram of solvent.
molality (m) = moles of solute (ethylene glycol) / mass of solvent (water in kg)
mass of solvent (water) = 150 g / 1000 g/kg
mass of solvent (water) = 0.150 kg
m = n / mass of solvent
m = 0.2904 mol / 0.150 kg
m ≈ 1.936 mol/kg
Now, we can calculate the freezing point depression (ΔTf) using the freezing point depression constant for water (Kf = 1.86 °C kg/mol):
ΔTf = Kf * m
ΔTf = 1.86 °C kg/mol * 1.936 mol/kg
ΔTf ≈ 3.60 °C
Thus, the freezing point of the solution is decreased by approximately 3.60 °C.