You're asked to prepare 125mL of .0321 M AgNO3. How many grams would you need of a smaple known to be 99.81% AgNO3 by mass?
masspureAgNO3=molmassAgNO3*volumeinLiter*molarity
MassAgNO3= massabove/.9981
Well, well, well, it seems like we have a chemistry question here. Let's calculate this step by step in a fun and clownish way!
To find out how many grams of AgNO3 we need, we first need to convert the volume (125 mL) into moles. But hold your horses! We have a concentration of 0.0321 M, which tells us the number of moles in 1 liter (1000 mL). So, we'll begin by converting our given volume to liters.
125 mL is equal to 0.125 liters (did you see how I did that? I just moved that pesky decimal point!).
Now that we have the volume in liters, let's multiply it by the molarity to get the number of moles. 0.125 liters multiplied by 0.0321 M gives us 0.0040125 moles of AgNO3 needed.
Now, let's unleash the clown magic! We know that our sample is 99.81% pure AgNO3. So, we need to find out how many grams that 0.0040125 moles of AgNO3 corresponds to.
Since the sample is 99.81% pure, we can multiply the moles by 0.9981 (99.81% in decimal form) to find the moles of AgNO3 in the sample.
0.0040125 moles multiplied by 0.9981 gives us 0.0039990375 moles of AgNO3.
Lastly, to find the grams, we need to multiply the moles by the molar mass of AgNO3, which is approximately 169.87 g/mol.
0.0039990375 multiplied by 169.87 gives us approximately 0.679 grams of AgNO3.
So, to prepare that jazzy 125 mL of 0.0321 M AgNO3 solution, you would need about 0.679 grams of your 99.81% pure sample!
As a clown, I must admit, those calculations were quite amusing!
To determine the amount of AgNO3 needed to prepare a 125mL solution with a concentration of 0.0321 M, we first need to calculate the number of moles of AgNO3 required.
The molar concentration (M) is defined as the number of moles of solute per liter of solution. Given that the desired concentration is 0.0321 M and the volume is 125mL (0.125L), we can calculate the moles of AgNO3 needed:
Moles of AgNO3 = Molarity x Volume
= 0.0321 mol/L x 0.125 L
= 0.00401 moles
To convert moles to grams, we need to use the molar mass of AgNO3. The molar mass of AgNO3 is calculated by summing the atomic masses of silver (Ag), nitrogen (N), and three oxygen (O) atoms:
Molar mass of AgNO3 = (atomic mass of Ag) + (atomic mass of N) + 3 x (atomic mass of O)
= (107.87 g/mol) + (14.01 g/mol) + 3 x (16.00 g/mol)
= 169.88 g/mol
To find the mass of AgNO3 needed, we can multiply the number of moles by the molar mass:
Mass of AgNO3 = Moles x Molar mass
= 0.00401 moles x 169.88 g/mol
= 0.6796 grams
However, the sample is reported to be 99.81% pure AgNO3. Therefore, we must consider this purity when calculating the mass needed. To do this, we divide the desired mass by the purity percentage:
Actual mass of sample = Mass of AgNO3 needed / Purity percentage
= 0.6796 g / 0.9981
= 0.6809 grams (rounded to four decimal places)
Therefore, you would need approximately 0.6809 grams of the AgNO3 sample that is known to be 99.81% pure by mass.
To find the number of grams needed, we can use the formula:
grams = volume (in liters) × molarity × molar mass
First, let's calculate the volume in liters:
125 mL = 125/1000 = 0.125 L
Next, we need to determine the molar mass of AgNO3 (silver nitrate):
AgNO3:
- Ag (silver) has a molar mass of 107.87 g/mol
- N (nitrogen) has a molar mass of 14.01 g/mol
- O (oxygen) has a molar mass of 16.00 g/mol (x3 since there are three oxygen atoms in AgNO3)
Molar mass of AgNO3 = 107.87 + 14.01 + (16.00 × 3) = 107.87 + 14.01 + 48.00 = 169.88 g/mol
Now we can substitute the values into the formula:
grams = 0.125 L × 0.0321 mol/L × 169.88 g/mol
Calculating this expression will give us the answer:
grams = 0.125 × 0.0321 × 169.88 ≈ 0.6865 g
Therefore, you would need approximately 0.6865 grams of the sample known to be 99.81% AgNO3 by mass.