You have 250 mL of a 0.80 M solution of silver nitrate AgNO3 (169.87 g/mol) dissolved in water. How many grams of AgNO3 does the solution contain?
How many moles AgNO3 do you have? That is M x L = moles.
Then moles = grams/molar mass. You know moles and molar mass; solve for grams.
sol.. ist you converted 250 ml in to liter, so it will be 0.25 l
then (0.25l)(0.8 M)(169.87 g/mol)
it will be 33.94 gram
s0 33.94 g Ag-nitrate require to make 250 ml of 0.8 M sol of silver nitrate
Or u can say that 250 ml of 0.8 M sol of silver nitrate contain 33.94 g Ag-nitrate.
To find the amount of AgNO3 in the solution, we can use the formula:
Amount of solute (in moles) = concentration (in M) * volume of solution (in L)
First, let's convert the volume of the solution from milliliters (mL) to liters (L):
250 mL = 250/1000 L
= 0.25 L
Now, using the formula mentioned above, we can calculate the amount of AgNO3 in moles:
Amount of AgNO3 (in moles) = 0.80 M * 0.25 L
= 0.20 moles
Next, we can calculate the mass of AgNO3 in grams using the molar mass of AgNO3:
Mass = number of moles * molar mass
Molar mass of AgNO3 = 169.87 g/mol
Mass of AgNO3 = 0.20 moles * 169.87 g/mol
= 33.974 g
Therefore, the solution contains approximately 33.974 grams of AgNO3.
To find the number of grams of AgNO3 in the solution, we need to use the following formula:
grams of AgNO3 = (volume of solution in L) x (molarity of solution in mol/L) x (molar mass of AgNO3 in g/mol)
First, let's convert the volume of the solution from milliliters (mL) to liters (L):
250 mL = 250/1000 L = 0.25 L
Next, we can substitute the values into the formula:
grams of AgNO3 = (0.25 L) x (0.80 mol/L) x (169.87 g/mol)
Now, we can perform the calculation:
grams of AgNO3 = 0.25 x 0.80 x 169.87 = 34.47 g
Therefore, the solution contains approximately 34.47 grams of AgNO3.