How many moles of each of the following are needed to give the same freezing point lowering as 5.0mol of the nonelectrolyte ethylene glycol in 1 kg of water?CH3−OH, a non electrolyte, KNO3, a strong electrolyte.
Im confused on what they are asking, and how in the world would you set this up
I believe you are over thinking the problem.
delta T for ethylene glycol is
dT = Kf*m. You can calculate m, then dT.
Substitute that dT and Kf and solve for m of CH3OH. Then m = mols/kg; you know kg and m, solve for mols. I think that is 5 mols CH3OH.
The reason for the problem is to recognize that KNO3 is not a non-electrolyte; it dissolves into two particles and you must use the version of the formula that is
dT = i*Kf*m where i is the van't Hoff factor which is 1(for non-electrolytes like ethylene glycol and methyl alcohol (CH3OH) but 2 for KNO3. It would be 3 for Na2SO4 and 2 for NaCl, etc.
So set up dT and i and Kf and solve for m and mols. I think that answer will be 2.5 mols needed although I didn't put a pen and paper to it and it's possible I goofed in my memory banks.
To determine the number of moles needed to produce the same freezing point lowering, you can use the equation for freezing point depression:
ΔT = Kf * molality * i
Where:
ΔT = freezing point depression
Kf = cryoscopic constant (given constant for water)
molality = moles of solute per kilogram of solvent
i = vant Hoff factor (number of ions produced by the solute)
In this problem, you need to find the number of moles of each solute required to produce the same freezing point depression as 5.0 mol of ethylene glycol.
First, let's calculate the molality of ethylene glycol in 1 kg of water. The molality (m) is defined as the moles of solute per kilogram of solvent:
m = (moles of solute) / (mass of solvent in kg)
We can rearrange the equation to find the moles of solute:
moles of solute = (molality) * (mass of solvent in kg)
Since the mass of solvent is given as 1 kg, the moles of ethylene glycol in 1 kg of water will also be 5.0 mol.
Now, let's calculate the moles of CH3−OH (methanol) and KNO3 needed to produce the same freezing point depression.
For CH3−OH, since it is a non-electrolyte, the vant Hoff factor (i) is 1. Therefore, we can use the same equation as above:
moles of CH3−OH = (molality of ethylene glycol) * (mass of solvent in kg)
moles of CH3−OH = 5.0 mol
For KNO3, it is a strong electrolyte and dissociates into two ions (K+ and NO3-) when dissolved in water. Thus, the vant Hoff factor (i) is 2. To find the moles of KNO3 needed, we can adjust the equation above:
moles of KNO3 = (molality of ethylene glycol) * (mass of solvent in kg) * (i)
moles of KNO3 = 5.0 mol * 2
moles of KNO3 = 10.0 mol
Therefore, to achieve the same freezing point depression as 5.0 mol of ethylene glycol in 1 kg of water, you would need 5.0 mol of CH3−OH or 10.0 mol of KNO3.
To answer the question, we need to use the concept of freezing point depression, which involves the lowering of the freezing point of a solvent due to the presence of a solute. The amount of freezing point depression is directly proportional to the number of moles of solute particles dissolved in the solvent.
In this case, we are given that 5.0 moles of the nonelectrolyte ethylene glycol (C2H6O2) in 1 kg of water produces a certain amount of freezing point depression. We want to determine how many moles of each of the given substances will produce the same freezing point depression.
Let's break down the problem step by step:
Step 1: Calculate the freezing point depression of the solution.
The freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m,
where Kf is the freezing point depression constant for the solvent (water) and m is the molality of the solute.
The freezing point depression constant (Kf) for water is known to be 1.86 °C·kg/mol.
Step 2: Calculate the molality (m) of ethylene glycol in the original solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
m = moles of solute / mass of solvent (in kg).
Given that the mass of water is 1 kg and the number of moles of ethylene glycol is 5.0 mol, we can calculate the molality (m) as follows:
m = 5.0 mol / 1 kg = 5.0 mol/kg.
Step 3: Calculate the number of moles of each solute needed to produce the same freezing point depression.
Since we know the molality, we can use the equation:
m = (moles of solute) / (mass of solvent in kg),
to solve for the number of moles of solute.
For CH3−OH (methanol), which is a non-electrolyte, the number of moles required can be determined as:
moles of CH3−OH = m * mass of solvent (in kg).
Similarly, for KNO3 (potassium nitrate), which is a strong electrolyte, the number of moles required can be determined as:
moles of KNO3 = (m * mass of solvent) / number of particles in one formula unit of KNO3.
It's important to note that KNO3 dissociates into two ions (K+ and NO3-) when dissolved in water. Therefore, the number of particles in one formula unit of KNO3 is 2.
By using these equations, you can calculate the number of moles of CH3−OH and KNO3 needed to give the same freezing point lowering as 5.0 moles of ethylene glycol in 1 kg of water.