A solution is prepared by dissolving 0.2222 (¡¾0.0002) g of KIO3 in 50.00 (¡¾0.05) mL.

(i) Calculate the molar mass of KIO3 with its associated uncertainty to the correct number of significant figures.
(ii) Find the molarity and its uncertainty with an appropriate number of significant figures.
(iii) Would the answer be affected significantly if the reagent were only 99.9% pure? Why or why not

I don't get this.... plz help me omg

No problem, I can help you! Let's go step by step to solve this problem.

(i) To find the molar mass of KIO3 with its associated uncertainty, we need to consider the atomic masses of potassium (K), iodine (I), and oxygen (O). The molar mass of KIO3 can be calculated by adding the atomic masses together.
- Atomic mass of K: 39.10 g/mol
- Atomic mass of I: 126.90 g/mol
- Atomic mass of O: 16.00 g/mol

Now, we can calculate the molar mass of KIO3:
Molar mass of KIO3 = (Atomic mass of K) + (Atomic mass of I) + 3 x (Atomic mass of O)
= 39.10 g/mol + 126.90 g/mol + 3 x 16.00 g/mol
= 214.10 g/mol

To find the associated uncertainty, we need to consider the given uncertainty in the mass of KIO3 (±0.0002 g). The molar mass uncertainty will be equal to the uncertainty in the mass.
Molar mass uncertainty = ±0.0002 g

Therefore, the molar mass of KIO3 with its associated uncertainty is 214.10 (±0.0002) g/mol.

(ii) To find the molarity (molar concentration) and its uncertainty, we need to use the given mass of KIO3 (0.2222 g ±0.0002 g) and volume (50.00 mL ±0.05 mL). However, we need to convert the volume from milliliters to liters:
Volume = 50.00 mL = 0.05000 L (since 1 mL = 0.001 L)

Now, we can calculate the molarity using the formula:
Molarity (M) = (Mass of solute in grams) / (Molar mass in grams/mol) / (Volume of solution in liters)

Molarity = (0.2222 g ±0.0002 g) / (214.10 g/mol ±0.0002 g/mol) / (0.05000 L ±0.00005 L)

To find the uncertainty in the molarity, we need to consider the uncertainties in the mass, molar mass, and volume. We can use the formula for error propagation to calculate it.

Molarity uncertainty = Molarity x √((Mass uncertainty / Mass)^2 + (Molar mass uncertainty / Molar mass)^2 + (Volume uncertainty / Volume)^2)

Plugging in the values and calculating the uncertainty, we get:

Molarity uncertainty = (Molarity) x √((0.0002 g / 0.2222 g)^2 + (0.0002 g/mol / 214.10 g/mol)^2 + (0.00005 L / 0.05000 L)²)

Now, let's calculate the molarity and its uncertainty:

Molarity = (0.2222 g) / (214.10 g/mol) / (0.05000 L) = 0.5220 mol/L

Molarity uncertainty = (0.5220 mol/L) x √((0.0002 g / 0.2222 g)^2 + (0.0002 g/mol / 214.10 g/mol)^2 + (0.00005 L / 0.05000 L)²)

After calculating the uncertainty, we get:

Molarity uncertainty = ±0.0005 mol/L

Therefore, the molarity of the solution is 0.5220 mol/L with an uncertainty of ±0.0005 mol/L.

(iii) If the reagent were only 99.9% pure, it means that 0.1% of the substance is impure. This impurity would affect the accuracy of the molarity calculation because the molar mass used in the calculation is based on 100% purity.

The molar mass calculation assumed that the KIO3 sample was 100% pure, but in reality, it is 99.9% pure. This means that there could be impurities present in the sample, causing the molar mass to deviate from its theoretical value. As a result, the calculated molarity may be slightly inaccurate.

In practical terms, if the impurity is significant enough, it would affect the molar mass and therefore the molarity calculation. However, in most cases, a 99.9% purity is still considered high and the effect on the molarity calculation would be negligible or insignificant.

Overall, if the reagent were only 99.9% pure, it would likely have a minimal impact on the calculated molarity and the answer would not be significantly affected.

No problem, I'll break it down step by step for you!

(i) To calculate the molar mass of KIO3, we need to know the atomic masses of the individual elements involved: potassium (K), iodine (I), and oxygen (O). Looking up their atomic masses, we find:

- Potassium (K): 39.10 g/mol
- Iodine (I): 126.90 g/mol
- Oxygen (O): 16.00 g/mol

Now we can calculate the molar mass of KIO3:
Molar Mass = (Atomic Mass of K) + (Atomic Mass of I) + 3 * (Atomic Mass of O)

Molar Mass = (39.10 g/mol) + (126.90 g/mol) + 3 * (16.00 g/mol)
Molar Mass = 214.10 g/mol

The molar mass of KIO3 is 214.10 g/mol.

To find the uncertainty in the molar mass, we consider the uncertainties in the atomic masses of the elements involved. The given uncertainty for the mass of KIO3 is 0.0002 g.

(ii) Now let's calculate the molarity of the solution. Molarity (M) is defined as the number of moles of solute (in this case, KIO3) divided by the volume of the solution in liters (L).

First, we need to find the number of moles of KIO3. We can use the formula:

Moles = Mass / Molar Mass

Moles = 0.2222 g / 214.10 g/mol
Moles = 0.001038 mol

Next, let's convert the given volume of the solution (50.00 mL) to liters:

Volume = 50.00 mL = 50.00 g / 1000 = 0.05000 L

Now we can calculate the molarity:
Molarity = Moles / Volume

Molarity = 0.001038 mol / 0.05000 L
Molarity = 0.02076 M

The molarity of the solution is 0.02076 M.

To find the uncertainty in the molarity, we need to consider the uncertainties in the mass of KIO3 and the volume of the solution. The given uncertainty for the mass of KIO3 is 0.0002 g, and the given uncertainty for the volume is 0.05 mL. We need to convert the volume uncertainty to liters:

Volume uncertainty = 0.05 mL = 0.05 g / 1000 = 0.00005 L

Now we can calculate the molarity uncertainty:
Molarity uncertainty = (Mass uncertainty / Molar Mass) + (Volume uncertainty / Volume^2)

Molarity uncertainty = (0.0002 g / 214.10 g/mol) + (0.00005 L / (0.05000 L)^2)
Molarity uncertainty = 9.338 x 10^-6 + 2.0 x 10^-7

The molarity uncertainty is 9.538 x 10^-6 M.

(iii) Finally, let's consider the effect of purity on the final answer. If the reagent (KIO3) is only 99.9% pure, it means that only 99.9% of the mass is due to KIO3, and the rest could be impurities.

Let's calculate the amount of actual KIO3 in the solution given its purity:

Actual KIO3 mass = (Mass of KIO3) * (Purity/100)

Actual KIO3 mass = 0.2222 g * (99.9/100)
Actual KIO3 mass = 0.221978 g

Using this actual KIO3 mass, we can recalculate the molarity:

Moles = Actual KIO3 mass / Molar Mass

Moles = 0.221978 g / 214.10 g/mol
Moles = 0.001035 mol

Molarity = Moles / Volume

Molarity = 0.001035 mol / 0.05000 L
Molarity = 0.02070 M

Comparing this result to the previous molarity calculation (0.02076 M), we can see that there is a slight difference due to impurities in the reagent. However, given that the difference is only in the fourth decimal place, it is unlikely to significantly affect the overall outcome.

I hope that helps!