A student dissolved 6.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3 - ion are there in the final solution?

How many grams of NO3^- are in the 100 mL. That's

6.00 x (2 x molar mass NO3^-/molar mass Co(NO3)2 = ? which I will call G.
Now, how many grams will be in the 4.00 mL? That's G x (4.00 mL/100 mL) = ? grams which is in the 275 mL.

Why did the Co(NO3)2 go to therapy? Because it felt like its ions were dissolving away.

Let's work it out! To find the grams of NO3- ion in the final solution, we need to look at the concentration of the stock solution and the dilution.

First, let's find the concentration of the stock solution:
Mass of Co(NO3)2 = 6.00 g
Volume of stock solution = 100. mL

So, the concentration of Co(NO3)2 in the stock solution is:
Concentration = (mass of Co(NO3)2) / (volume of stock solution)
Concentration = 6.00 g / 100. mL

Now, let's find the amount of Co(NO3)2 in the 4.00 mL of stock solution taken:
Amount of Co(NO3)2 = (concentration of stock solution) x (volume of stock solution taken)
Amount of Co(NO3)2 = (6.00 g / 100. mL) x 4.00 mL

Next, let's find the concentration of the final solution:
Volume of final solution = 275. mL

Since the stock solution was diluted in a ratio of 1:275/4, the concentration of Co(NO3)2 in the final solution is:
Concentration = (concentration of stock solution) x (volume of stock solution taken) / (volume of final solution)
Concentration = [(6.00 g / 100. mL) x 4.00 mL] / 275. mL

Finally, let's find the grams of NO3- ion in the final solution by multiplying the concentration by the molar mass of NO3-:
Grams of NO3- ion = (concentration in final solution) x (molar mass of NO3-)

I hope that cleared things up for you! Who knew there could be so much chemistry in a clown's toolbox?

To solve this problem, we need to calculate the concentration of the Co(NO3)2 stock solution and then use that information to find the amount of NO3- ions in the final solution. Here are the steps:

Step 1: Calculate the molarity of the Co(NO3)2 stock solution.
Molarity (M) = moles of solute / liters of solution

First, calculate the moles of Co(NO3)2:
Molar mass of Co(NO3)2 = 58.93 g/mol + 2 * (14.01 g/mol + 3 * 16.00 g/mol) = 165.93 g/mol
Moles of Co(NO3)2 = mass of Co(NO3)2 / molar mass of Co(NO3)2
Moles of Co(NO3)2 = 6.00 g / 165.93 g/mol ≈ 0.0361 mol

Next, calculate the volume of the stock solution in liters:
Volume of stock solution = 100. mL / 1000 mL/L = 0.100 L

Now, we can calculate the molarity:
Molarity of Co(NO3)2 stock solution = moles of Co(NO3)2 / volume of stock solution
Molarity of Co(NO3)2 stock solution = 0.0361 mol / 0.100 L = 0.361 M

Step 2: Calculate the moles of Co(NO3)2 in the final solution.
Moles of Co(NO3)2 in final solution = molarity of Co(NO3)2 stock solution * volume of stock solution in liters
Moles of Co(NO3)2 in final solution = 0.361 M * 4.00 mL / 1000 mL/L = 0.00144 mol

Step 3: Calculate the moles of NO3- ions in the final solution.
From the chemical formula, we can see that there are 2 moles of NO3- ions for every 1 mole of Co(NO3)2.
Moles of NO3- ions in final solution = 2 * moles of Co(NO3)2 in final solution = 2 * 0.00144 mol = 0.00288 mol

Step 4: Calculate the mass of NO3- ions.
Mass of NO3- ions = moles of NO3- ions * molar mass of NO3-
The molar mass of NO3- is 14.01 g/mol + 3 * 16.00 g/mol = 62.01 g/mol.
Mass of NO3- ions = 0.00288 mol * 62.01 g/mol = 0.178 g

Therefore, there are approximately 0.178 grams of NO3- ions in the final solution.

To find the number of grams of NO3- ions in the final solution, we need to consider the dilution factor and convert the given information into appropriate units for calculations.

Let's break down the problem into steps:

Step 1: Calculate the concentration of the stock solution
The student dissolved 6.00 g of Co(NO3)2 in enough water to make 100 mL of stock solution. We can calculate the concentration (C1) of the stock solution using the formula:

C1 = (mass of solute) / (volume of solution)

C1 = 6.00 g / 100. mL

C1 = 0.06 g/mL

Step 2: Calculate the amount of Co(NO3)2 in the volume taken from the stock solution
The student took 4.00 mL of the stock solution. We can calculate the mass (m1) of Co(NO3)2 in this volume using the formula:

m1 = C1 * V1

m1 = 0.06 g/mL * 4.00 mL

m1 = 0.24 g

Step 3: Calculate the concentration of the final solution
The final solution is obtained by diluting the 4.00 mL of stock solution to a total volume of 275 mL. We can calculate the concentration (C2) of the final solution using the formula:

C2 = (C1 * V1) / V2

C2 = (0.06 g/mL * 4.00 mL) / 275 mL

C2 = 0.00087272 g/mL

Step 4: Calculate the amount of NO3- ions in the final solution
Since Co(NO3)2 contains 2 NO3- ions for every 1 Co2+ ion, the amount of NO3- ions (m2) in the final solution is twice the mass of Co(NO3)2 (m1):

m2 = 2 * m1

m2 = 2 * 0.24 g

m2 = 0.48 g

Therefore, there are 0.48 grams of NO3- ions in the final solution.