What mass of silver nitrate is dissolved in 40.0 mL of a 0.400 M solution of silver nitrate?
40 ml of 0.4M solution contains 0.04*0.4
=0.016 mole of AgNO3.
RMM of AgNO3 = 107.87+14+3*16=169.87
So 0.016*169.87=2.72 g
Well, let's think about it. If there's already a solution of silver nitrate with a concentration of 0.400 M, then it means that 1 liter of that solution contains 0.400 moles of silver nitrate. But we only have 40.0 mL of the solution, which is 0.040 L. So, we can find the number of moles of silver nitrate in this case by multiplying the volume (in liters) by the concentration (in moles per liter).
0.040 L x 0.400 M = 0.016 moles of silver nitrate
Now, to find the mass of silver nitrate, we need to know its molar mass. The molar mass of silver nitrate (AgNO3) is approximately 169.87 g/mol. So, let's multiply the number of moles by the molar mass:
0.016 moles x 169.87 g/mol = 2.718 grams (rounded to three decimal places)
Therefore, the mass of silver nitrate dissolved in 40.0 mL of the 0.400 M solution is approximately 2.718 grams. Just remember, math may not always be funny, but it still gets the job done!
To find the mass of silver nitrate dissolved in the solution, we need to use the formula:
Mass = Volume x Concentration
Given:
Volume (V) = 40.0 mL = 0.0400 L
Concentration (C) = 0.400 M
Substituting the values into the formula:
Mass = 0.0400 L x 0.400 M
Mass = 0.0160 moles
Since we now have the number of moles of silver nitrate, we can convert it to mass using the molar mass of silver nitrate (AgNO3):
Molar mass of AgNO3 = 107.87 g/mol (for Ag) + 14.01 g/mol (for N) + (3 x 16.00 g/mol) (for O)
Molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + 48.00 g/mol
Molar mass of AgNO3 = 169.88 g/mol
Now, we can calculate the mass of silver nitrate:
Mass = 0.0160 moles x 169.88 g/mol
Mass ≈ 2.718 g
Therefore, approximately 2.718 grams of silver nitrate is dissolved in 40.0 mL of a 0.400 M solution of silver nitrate.
To find the mass of silver nitrate dissolved in the solution, you need to use the formula:
mass = volume × concentration × molar mass
First, let's calculate the number of moles of silver nitrate using the given values:
volume = 40.0 mL (convert to L by dividing by 1000)
concentration = 0.400 M
Since concentration is in moles per liter (M), we need to convert the volume to liters:
40.0 mL ÷ 1000 = 0.0400 L
Now, we can calculate the number of moles of silver nitrate:
moles = volume × concentration
moles = 0.0400 L × 0.400 M
Next, we need the molar mass of silver nitrate (AgNO3). The atomic masses of each element are:
Ag (silver) = 107.87 g/mol
N (nitrogen) = 14.01 g/mol
O (oxygen) = 16.00 g/mol (there are three oxygen atoms in silver nitrate)
molar mass of AgNO3 = Ag + N + (O × 3)
molar mass of AgNO3 = 107.87 g/mol + 14.01 g/mol + (16.00 g/mol × 3)
Finally, we can calculate the mass of silver nitrate:
mass = moles × molar mass
Substitute the calculated values into the equation to get the result.