The vapor pressure of pure water at 85 celsius is normally 433.6 mmHg but decrease to 393.8mmHg upon addition?

of an unknown amount of ammoniu chloride NH4Cl to 162.2g of water at this temperature. How many grams of ammonium chloride were added assuming complete dissociation of solute

delta P = XNH4Cl*Po

433.6-393.8 = XNH4Cl*433.6
XNH4Cl = 39.8/433.6 = ??
Then n = moles NH4Cl
162.2/18.015 = 9.00 moles H2O
2n/(2n+9) = 39.8/433.6 and solve for n.
Then grams NH4Cl = n NH4Cl * molar mass NH4Cl.

Well, I see we have some molecular shenanigans going on here! Let's calculate this with a touch of humor.

To figure out how much ammonium chloride was added, we first need to find the decrease in vapor pressure. That's like trying to find the lost sock in the laundry!

We start with the given, the vapor pressure of pure water is 433.6 mmHg. Suddenly, it drops to 393.8 mmHg. Oh no, it must be feeling a bit shy or maybe it's just having a bad hair day!

So, the decrease in vapor pressure is 433.6 mmHg - 393.8 mmHg = 39.8 mmHg. It's like the water is wearing a disguise!

Now, the decrease in vapor pressure is directly proportional to the concentration of the solute. If we assume complete dissociation (which is like assuming there are no backup dancers for the water molecule), we can relate this change to the mass of ammonium chloride added.

The molecular weight of ammonium chloride (NH4Cl) is 53.49 g/mol. So, for every mole of ammonium chloride added, we get a decrease of 39.8 mmHg in vapor pressure.

Now comes the fun part! Let's use the given mass of water (162.2 g) to calculate the amount of ammonium chloride added, assuming it dissociates completely. It's like playing a game of hide and seek with the unknown!

First, let's convert the mass of water to moles. The molar mass of water (H2O) is approximately 18.02 g/mol. So, 162.2 g of water is around 9 moles.

Since we now know that for every mole of ammonium chloride we get a decrease in vapor pressure of 39.8 mmHg, we can set up a proportion to find the moles of ammonium chloride added. It's like setting up a fair trade agreement!

9 moles of water / x moles of ammonium chloride = 39.8 mmHg / 1 mole of ammonium chloride.

Now, cross-multiply and solve for x:

9 moles of water * 1 mole of ammonium chloride = x moles of ammonium chloride * 39.8 mmHg.

x = (9 moles of water * 1 mole of ammonium chloride) / 39.8 mmHg.

Calculating this, we find x to be approximately 0.226 moles of ammonium chloride. It's like finding a little circus performer in a big box!

Finally, let's convert moles to grams. One mole of ammonium chloride (NH4Cl) has a mass of 53.49 g. So, the mass of ammonium chloride added is approximately 0.226 moles * 53.49 g/mol = 12.10 g.

Tada! Approximately 12.10 grams of ammonium chloride were added. It's like adding a pinch of spice to your culinary adventures!

To find the amount of ammonium chloride (NH4Cl) added assuming complete dissociation of the solute, we can use Raoult's law, which states that the vapor pressure of the solvent above the solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Let's calculate the mole fraction of water (solvent) in the solution:
Given:
- Vapor pressure of pure water at 85°C = 433.6 mmHg
- Vapor pressure of the solution with ammonium chloride at 85°C = 393.8 mmHg

Using Raoult's law, we can write the equation as:
393.8 mmHg = Xwater * 433.6 mmHg

To find Xwater (mole fraction of water), divide both sides of the equation by 433.6 mmHg:
Xwater = 393.8 mmHg / 433.6 mmHg
Xwater ≈ 0.909

The mole fraction of water is approximately 0.909.

Now, let's find the mole fraction of ammonium chloride (NH4Cl) in the solution:
Since ammonium chloride dissociates completely in water, the mole fraction of NH4Cl will be equal to the mole fraction of water.

XNH4Cl ≈ Xwater ≈ 0.909

We are given the mass of water in the solution as 162.2 g.
To find the number of moles of water, divide the mass by the molar mass of water.
Molar mass of water (H2O) = 18.015 g/mol
Number of moles of water = 162.2 g / 18.015 g/mol ≈ 9.00 mol

Since the mole fraction is equal to the number of moles of the component divided by the total moles of all components, we can write the equation:
XNH4Cl = Moles of NH4Cl / (Moles of NH4Cl + Moles of H2O)

Substituting the values:
0.909 = Moles of NH4Cl / (Moles of NH4Cl + 9.00 mol)

Now we can solve for the moles of NH4Cl:
0.909 * (Moles of NH4Cl + 9.00 mol) = Moles of NH4Cl
Moles of NH4Cl + 8.181 mol = 0.909 * Moles of NH4Cl
0.091 * Moles of NH4Cl = 8.181 mol
Moles of NH4Cl ≈ 8.181 mol / 0.091 ≈ 89.989 mol

The number of moles of ammonium chloride is approximately 89.989 mol.
To find the mass of ammonium chloride, multiply the number of moles by the molar mass of NH4Cl:
Molar mass of ammonium chloride (NH4Cl) = 53.49 g/mol (approx.)
Mass of NH4Cl = 89.989 mol * 53.49 g/mol ≈ 4812.31 g

Therefore, approximately 4812.31 grams of ammonium chloride were added assuming complete dissociation of the solute.

To determine the amount of ammonium chloride (NH4Cl) added to 162.2g of water, we can use the Raoult's law equation for calculating vapor pressure.

Raoult's law states that the vapor pressure of a solvent over a solution is directly proportional to the mole fraction of the solvent. In this case, the solvent is water (H2O) and the solute is ammonium chloride (NH4Cl).

The equation for Raoult's law is:
P = X_solvent * P_solvent

Where:
P is the vapor pressure of the solution.
X_solvent is the mole fraction of the solvent.
P_solvent is the vapor pressure of the pure solvent.

We have the vapor pressure of pure water at 85°C, which is 433.6 mmHg. And we know that the vapor pressure decreases to 393.8 mmHg upon addition of ammonium chloride.

First, let's calculate the mole fraction of the solvent (water).

Mole fraction (X_solvent) = moles of solvent / total moles of solution

The total moles of the solution is equal to the moles of water (given by its mass divided by its molar mass) plus the moles of ammonium chloride (which we need to find).

Molar mass of water (H2O) = 18.015 g/mol
Molar mass of ammonium chloride (NH4Cl) = 53.49 g/mol

Now let's calculate the moles of water:

moles of water = mass of water / molar mass of water
= 162.2 g / 18.015 g/mol
= 9.001 mol

Since ammonium chloride completely dissociates in water, it breaks up into one NH4+ ion and one Cl- ion.

To find the moles of ammonium chloride, we can equate the decrease in vapor pressure to the mole fraction of water:

P_solvent - P = X_solvent * P_solvent

393.8 mmHg - 433.6 mmHg = (moles of water) / (moles of water + moles of NH4Cl) * 433.6 mmHg

-39.8 mmHg = (9.001 mol) / (9.001 mol + moles of NH4Cl) * 433.6 mmHg

Simplifying the above equation, we get:

-39.8 mmHg = 9.001 mol / (9.001 mol + moles of NH4Cl) * 433.6 mmHg

Now, let's solve for the moles of ammonium chloride (NH4Cl). Rearranging the equation:

(-39.8 mmHg / 433.6 mmHg) * (9.001 mol + moles of NH4Cl) = 9.001 mol

Solving for moles of NH4Cl:

(-39.8 mmHg / 433.6 mmHg) * 9.001 mol = 9.001 mol + moles of NH4Cl

(-39.8 mmHg / 433.6 mmHg) * 9.001 mol - 9.001 mol = moles of NH4Cl

The moles of NH4Cl are the same as the moles of ammonium chloride since it dissociates completely in water.

Now we can convert the moles of NH4Cl to grams:

mass of NH4Cl = moles of NH4Cl * molar mass of NH4Cl

Substituting the values, we get:

mass of NH4Cl = moles of NH4Cl * 53.49 g/mol

Finally, calculate the mass of NH4Cl added by substituting the moles of NH4Cl:

mass of NH4Cl = (-39.8 mmHg / 433.6 mmHg) * 9.001 mol * 53.49 g/mol

Calculating the above expression will give you the mass of ammonium chloride added in grams.