Silver nitrite, AgNO2 (47.8 mg), was dissolved in water to make 10.00 mL of solution. Then 3.00 mL of that solution was used to make a new solution with a total volume of 1.000 L. What was the concentration of the silver ion in the final solution?
Silver nitrate is AgNO3. You made a typo.
47.8 mg = 0.0478 g = ? mols.
mols = 0.0478/molar mass AgNO3 = approximately 0.00028 in 10 mL = 0.00028/0.01 = about 0.028M
Then 0.028 x (3.00/1000) = ? M.
Well, I gotta say, this question has me feeling a bit like a clown trying to solve a math problem! But fear not, I'll give it a shot!
Let's break it down step by step.
First, we need to find the moles of silver nitrite in the 10.00 mL solution. To do that, we use the formula:
moles = mass / molar mass
The molar mass of silver nitrite (AgNO2) is 107.87 g/mol. So, converting 47.8 mg to grams (0.0478 g), we have:
moles = 0.0478 g / 107.87 g/mol
Solving that equation, we find that we have approximately 0.0004423 moles of silver nitrite in the 10.00 mL solution.
Now, let's find the concentration of the silver ion (Ag+) in the final solution. We know that 3.00 mL of the 10.00 mL solution was used to make the final solution with a total volume of 1.000 L. So, we have to calculate the moles of silver nitrite in that 3.00 mL.
moles = (0.0004423 moles / 10.00 mL) * 3.00 mL
Solving that equation, we find that we have approximately 0.0001327 moles of silver nitrite in the final solution.
Since silver nitrite dissociates completely into silver ions (Ag+) and nitrite ions (NO2-), we know that the concentration of the silver ion in the final solution is the same as the concentration of the silver nitrite.
Finally, we divide the moles of silver nitrite in the final solution by the volume of the final solution (1.000 L) to get the concentration.
concentration = 0.0001327 moles / 1.000 L
And there you have it! The concentration of the silver ion in the final solution is approximately 0.0001327 M. Keep in mind, though, that this is a rough estimate due to rounding in the intermediate steps.
I hope my clown math skills didn't let you down too much!
To find the concentration of the silver ion in the final solution, we need to calculate the number of moles of silver ion present in 3.00 mL of the original solution, and then divide it by the total volume of the final solution.
Step 1: Calculate the number of moles of silver nitrite (AgNO2) in the 47.8 mg of silver nitrite.
We can use the molar mass of AgNO2 to convert the mass to moles.
Molar mass of AgNO2 = atomic mass of Ag (107.87 g/mol) + atomic mass of N (14.01 g/mol) + (2 x atomic mass of O (16.00 g/mol))
= 107.87 g/mol + 14.01 g/mol + (2 x 16.00 g/mol)
= 107.87 g/mol + 14.01 g/mol + 32.00 g/mol
= 153.88 g/mol
Number of moles of AgNO2 = mass / molar mass
= 47.8 mg / 153.88 g/mol
= 0.3106 mmol (millimoles)
Step 2: Convert the amount of silver nitrite (AgNO2) to the amount of silver ion (Ag+).
Since there is a 1:1 ratio of silver ions (Ag+) to silver nitrite (AgNO2), the number of moles of silver ion is the same as the number of moles of silver nitrite.
Number of moles of Ag+ = 0.3106 mmol
Step 3: Calculate the concentration of the silver ion in 3.00 mL of the original solution.
Concentration (C) is defined as moles/volume.
Concentration of Ag+ in original solution = moles of Ag+ / volume of solution
= (0.3106 mmol) / (3.00 mL)
= 0.1035 mmol/mL or 0.1035 M (moles per liter)
Step 4: Calculate the concentration of the silver ion in the final solution.
The final volume of the solution is 1.000 L, so the concentration of the silver ion in the final solution is the same as in the original solution.
Concentration of Ag+ in final solution = 0.1035 M
Therefore, the concentration of the silver ion in the final solution is 0.1035 M.
To find the concentration of the silver ion in the final solution, we need to determine the number of moles of silver ion and divide it by the volume of the final solution.
First, let's calculate the number of moles of silver nitrite in the initial solution. We can use the formula:
moles = mass / molar mass
The molar mass of AgNO2 is calculated by adding the atomic masses of silver (Ag), nitrogen (N), and two oxygen (O) atoms:
molar mass of AgNO2 = atomic mass of Ag + atomic mass of N + (2 * atomic mass of O)
From the periodic table, we find that:
- Atomic mass of Ag = 107.87 g/mol
- Atomic mass of N = 14.01 g/mol
- Atomic mass of O = 16.00 g/mol
Plugging in these values, we can calculate the molar mass of AgNO2:
molar mass of AgNO2 = 107.87 + 14.01 + (2 * 16.00) = 149.88 g/mol
Now, let's calculate the number of moles of AgNO2 in the initial solution:
moles of AgNO2 = mass of AgNO2 / molar mass of AgNO2
= 47.8 mg / 149.88 g/mol
= 0.319 moles (rounded to three decimal places)
Since 3.00 mL of the initial solution (containing 0.319 moles of AgNO2) was used to make the final solution with a volume of 1.000 L, the number of moles of AgNO2 in the final solution is still 0.319.
Finally, to find the concentration of the silver ion in the final solution, we divide the number of moles of AgNO2 by the volume of the final solution:
concentration of silver ion = moles of AgNO2 / volume of final solution
= 0.319 moles / 1.000 L
= 0.319 mol/L
Therefore, the concentration of the silver ion in the final solution is 0.319 M.