You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.7 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.264 g. What was the concentration of the original lead(II) nitrate solution?

If the reaction is

Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3

then each mole of PbCl2 uses one mole of Pb(NO3)2.

So, with 3.264/278.11 = 0.0117 moles of PbCl2 produced, the 2.00mL was .0117/.002 = 5.868M

5.868*80.7/100 = 4.735M original concentration

Well, it looks like you've got a bit of a mystery on your hands! To figure out the concentration of the original lead(II) nitrate solution, we need to do a little detective work.

First, let's calculate the moles of solid lead chloride that formed when you mixed the lead(II) nitrate solution with the sodium chloride solution. We can use the mass of the solid and the molar mass of lead chloride.

So, the moles of lead chloride = mass / molar mass = 3.264 g / (molar mass of PbCl2)

But here's the catch, I don't actually know the molar mass of lead chloride off the top of my head. Hang on, let me consult the Molar Mass Dictionary...

(A few seconds pass...)

Ah-ha! According to my trustworthy Molar Mass Dictionary, the molar mass of lead chloride (PbCl2) is 278.10 g/mol.

So, the moles of lead chloride = 3.264 g / 278.10 g/mol

Now, the balanced chemical equation for the reaction between lead(II) nitrate and sodium chloride is:

Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3

From this equation, we can see that 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride to produce 1 mole of lead chloride. So the moles of lead(II) nitrate = the moles of lead chloride.

Well, this investigation is certainly turning into a chemistry mystery! But don't worry, we're getting close to solving it.

We know that the volume of the original solution was initially 100.0 mL but decreased to 80.7 mL due to evaporation. Now, we have 2.00 mL of the solution, which we used to react with the sodium chloride.

Using the volumes, we can set up a proportion to find the initial moles of lead(II) nitrate:

Initial moles / initial volume = final moles / final volume

Initial moles / 100.0 mL = moles of lead chloride / 2.00 mL

Solving for initial moles gives us:

Initial moles = (moles of lead chloride / 2.00 mL) * 100.0 mL

Now we just need to plug in the moles of lead chloride, which we calculated earlier, and calculate the initial moles of lead(II) nitrate.

Once we have the initial moles, we can calculate the concentration of the original lead(II) nitrate solution by dividing the initial moles by the volume (in liters) of the original solution, which is 0.100 L.

So, by taking all these calculated steps, you should be able to crack this chemistry mystery and determine the concentration of the original lead(II) nitrate solution!

Remember, being a detective can be challenging, but solving the case is always worth it in the end. Good luck, Sherlock!

To find the concentration of the original lead(II) nitrate solution, we can use the information provided.

First, we need to determine the initial number of moles of lead(II) nitrate in the solution before it evaporated. We can calculate this using the remaining volume and the molarity of the solution.

Given:
- Initial volume of the solution = 100.0 mL
- Final volume of the solution = 80.7 mL

The volume of lead(II) nitrate solution that evaporated can be calculated as:
Evaporated volume = Initial volume - Final volume
Evaporated volume = 100.0 mL - 80.7 mL
Evaporated volume = 19.3 mL

Next, we need to determine the initial amount of lead(II) nitrate in moles:
Initial moles of lead(II) nitrate = Initial volume x Initial concentration

Since we don't know the initial concentration, we will define it as "x mol/L."

Initial moles of lead(II) nitrate = (100.0 mL / 1000 mL) x x mol/L
Initial moles of lead(II) nitrate = 0.1 x (x mol/L)

The moles of lead(II) nitrate after the evaporation is given by:
Final moles of lead(II) nitrate = (80.7 mL / 1000 mL) x x mol/L
Final moles of lead(II) nitrate = 0.0807 x (x mol/L)

Since the moles of lead(II) nitrate are conserved, we can write the equation as:
Initial moles - Final moles = Mass of solid / Molar mass of lead(II) nitrate

0.1x - 0.0807x = 3.264 g / (207.2 g/mol)

Simplifying the equation:
0.0193x = 0.015753
x = 0.815 mol/L

Therefore, the concentration of the original lead(II) nitrate solution was 0.815 mol/L.

To find the concentration of the original lead(II) nitrate solution, we need to use the given information about the volume of the solution and the mass of the solid formed.

First, we need to determine the moles of lead(II) nitrate reacted with the sodium chloride.

Step 1: Convert the mass of the solid obtained to moles.
To do this, divide the mass of the solid (3.264 g) by the molar mass of the solid (which we need to determine).

The molar mass of the solid can be calculated using the chemical equation of the reaction between lead(II) nitrate (Pb(NO3)2) and sodium chloride (NaCl):
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3

The equation shows that 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride to produce 1 mole of lead(II) chloride.

Step 2: Calculate the molar mass of the solid.
To find the molar mass, we need to know the elemental composition of the solid. From the equation, we can determine that lead(II) chloride (PbCl2) is formed. So we need to find the molar mass of PbCl2.

The molar mass of Pb is 207.2 g/mol, and Cl has a molar mass of 35.5 g/mol. Since there are two Cl atoms in PbCl2, we multiply the molar mass of Cl by 2. Adding up the molar masses, we get:
207.2 g/mol + (35.5 g/mol × 2) = 278.2 g/mol

So the molar mass of the solid is 278.2 g/mol.

Step 3: Calculate the moles of the solid obtained.
Divide the mass of the solid obtained (3.264 g) by the molar mass of the solid (278.2 g/mol) to obtain the moles of the solid:
3.264 g ÷ 278.2 g/mol ≈ 0.0117 mol

Step 4: Determine the moles of lead(II) nitrate reacted.
Since the reaction between lead(II) nitrate and sodium chloride occurs in a 1:1 ratio, the moles of lead(II) nitrate reacted are also approximately 0.0117 mol.

Next, we need to find the initial moles of lead(II) nitrate in the 2.00 mL sample.

Step 5: Calculate the initial moles of lead(II) nitrate.
We know that the volume of the original lead(II) nitrate solution was 100.0 mL, but it evaporated down to 80.7 mL. Therefore, the ratio of the remaining volume to the initial volume is:
80.7 mL ÷ 100.0 mL = 0.807

Multiply this ratio by the moles of lead(II) nitrate reacted to find the initial moles:
0.807 × 0.0117 mol ≈ 0.00945 mol

Finally, we can determine the concentration of the original lead(II) nitrate solution.

Step 6: Calculate the concentration of the original solution.
The concentration is defined as moles of solute per volume of solution. In this case, the solute is lead(II) nitrate and the volume is 2.00 mL (the sample taken).

Since the sample is only a fraction of the original volume, we need to scale the initial moles of lead(II) nitrate to the original volume.

The original volume is 100.0 mL, so we multiply the initial moles by the ratio of the sample volume to the original volume:
(0.00945 mol ÷ 2.00 mL) × 100.0 mL = 0.4725 mol/L

Therefore, the concentration of the original lead(II) nitrate solution is approximately 0.4725 mol/L.