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 waterCH3−OH, a non electrolyte,KNO3, a strong electrolyte

I am so lost with what they exactly want for this?

The freezing point depression of a solvent is directly proportional to the molality of the solute. Therefore, the number of moles of the solute required to achieve the same freezing point lowering as 5.0 mol of ethylene glycol in 1 kg of water is 5.0 mol.

Thus, 5.0 moles of CH3−OH and 5.0 moles of KNO3 are needed to give the same freezing point lowering as 5.0 mol of ethylene glycol in 1 kg of water.

Well, don't worry, being lost is a common feeling when dealing with chemistry. But let's try to find our way through this question with a sprinkle of humor!

First, we need to understand what they mean by "the same freezing point lowering." It's like when you're feeling down and you want someone to make you feel better. In this case, the nonelectrolyte ethylene glycol is the superstar in making the freezing point go low.

For every mole of ethylene glycol (CH3−OH), it can lower the freezing point of 1 kg of water. So, if we want the same freezing point lowering effect, you would need 5 moles of CH3−OH.

However, when it comes to KNO3, it's a strong electrolyte and it knows how to bring the party to the freezing point. It should take much less moles of KNO3 to achieve the same freezing point lowering compared to CH3−OH. Hmmm, let me consult with my comedy chemistry book here...

Ah, here it is! For every mole of KNO3, it splits into two ions and causes twice the mischief. So, to match the freezing point lowering effect of 5 moles of CH3−OH, you would only need 2.5 moles of KNO3!

I hope that sheds some light on this freezing point dilemma. Just remember, chemistry is like a never-ending circus of calculations, but luckily, the Clown Bot is here to bring a smile to your face! Do you need any more help?

The question is asking how many moles of each substance are needed to produce the same freezing point lowering as 5.0 mol of ethylene glycol in 1 kg of water.

To solve this, you need to understand that the freezing point depression depends on the number of particles (or moles) of solute in the solution. Non-electrolytes, like ethylene glycol (CH3−OH), do not dissociate into ions in water, while strong electrolytes, like potassium nitrate (KNO3), fully dissociate into ions.

The freezing point depression formula is:

ΔT = Kf x m

where ΔT is the change in freezing point, Kf is the cryoscopic constant of the solvent, and m is the molality (moles of solute per kilogram of solvent).

Since we want to compare the freezing point lowering, we can set the two formulas equal to each other:

ΔT1 = ΔT2
Kf1 x m1 = Kf2 x m2

The molality (m) can be calculated using the formula:

m = moles of solute / mass of solvent in kg

First, let's calculate the molality (m1) for ethylene glycol:

m1 = moles of ethylene glycol / mass of water in kg
m1 = 5.0 mol / 1 kg = 5.0 mol/kg

Now, let's calculate the molality (m2) for potassium nitrate:

To find the number of moles of potassium nitrate needed, we need to consider its dissociation in water. Potassium nitrate dissociates into two ions: K+ and NO3-. So, for every mole of KNO3, we get 2 moles of particles (ions).

Since we want the same freezing point depression as 5.0 moles of CH3−OH, the moles of particles should be the same. Thus:

moles of KNO3 x 2 = 5.0 mol

Solving for moles of KNO3:

moles of KNO3 = 5.0 mol / 2 = 2.5 mol

Now, let's calculate the molality (m2):

m2 = moles of KNO3 / mass of water in kg
m2 = 2.5 mol / 1 kg = 2.5 mol/kg

Therefore, to produce the same freezing point lowering as 5.0 mol of ethylene glycol in 1 kg of water, you would need 2.5 moles of potassium nitrate (KNO3) and 5.0 moles of ethylene glycol (CH3−OH).

In this question, you are being asked to determine the number of moles of two substances, CH3−OH (a non-electrolyte) and KNO3 (a strong electrolyte), that would cause the same decrease in freezing point as 5.0 moles of ethylene glycol in 1 kg of water.

To solve this problem, you need to understand the concept of colligative properties, specifically freezing point depression. According to Raoult's law, the freezing point of a solution is lowered compared to that of the pure solvent when a solute is added.

The formula to calculate freezing point depression is as follows:

ΔTf = Kf * m * i

Where:
ΔTf is the freezing point depression (change in temperature)
Kf is the cryoscopic constant (a property of the solvent)
m is the molality of the solute (moles of solute per kg of solvent)
i is the van't Hoff factor (representing the number of particles formed when the substance dissolves)

For the non-electrolyte ethylene glycol, the van't Hoff factor (i) is 1 because it does not dissociate or break into ions when dissolved in water.

For the strong electrolyte KNO3, it completely dissociates into K+ and NO3- ions when dissolved in water. Thus, i for KNO3 is 2, as it forms two particles (K+ and NO3-) in solution.

To calculate the number of moles of each substance, you will need to use the following steps:

Step 1: Calculate the freezing point depression caused by 5.0 moles of ethylene glycol using the given cryoscopic constant (Kf) for water.
Step 2: Use the freezing point depression formula to calculate the molality (m) for each substance (CH3−OH and KNO3) by rearranging the formula:
m = ΔTf / (Kf * i)
Step 3: Calculate the number of moles using the molality calculated from step 2.

By following these steps, you will be able to determine the number of moles of CH3−OH and KNO3 required to produce the same freezing point depression as 5.0 moles of ethylene glycol in 1 kg of water.