Question
Calculate the volume in milliliters of a 3.39M silver nitrate solution that contains 75.0 g of silver nitrate (AgNO3) Be sure your answer has the correct number of significant digits.
Answers
GPT 3.5
To calculate the volume of the solution, first we need to find the molar mass of AgNO3:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol
Total molar mass = 107.87 + 14.01 + (3 * 16) = 169.88 g/mol
Next, we will use the formula for molarity:
Molarity = moles of solute / volume of solution in liters
Rearranging for volume of solution, we get:
Volume = moles of solute / molarity
First, we need to find the moles of AgNO3 in 75.0 g:
moles = mass / molar mass = 75.0 g / 169.88 g/mol = 0.4414 moles
Now we can calculate the volume of the solution:
Volume = 0.4414 moles / 3.39 mol/L = 0.130 L = 130 mL
Therefore, the volume of the 3.39M silver nitrate solution that contains 75.0 g of AgNO3 is 130 mL.
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol
Total molar mass = 107.87 + 14.01 + (3 * 16) = 169.88 g/mol
Next, we will use the formula for molarity:
Molarity = moles of solute / volume of solution in liters
Rearranging for volume of solution, we get:
Volume = moles of solute / molarity
First, we need to find the moles of AgNO3 in 75.0 g:
moles = mass / molar mass = 75.0 g / 169.88 g/mol = 0.4414 moles
Now we can calculate the volume of the solution:
Volume = 0.4414 moles / 3.39 mol/L = 0.130 L = 130 mL
Therefore, the volume of the 3.39M silver nitrate solution that contains 75.0 g of AgNO3 is 130 mL.
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