what mass of AgNO3(MW 169.87) is required to make 25 mL of 0.100 M AgNO3?
0.100 mole/liter * 0.025 liters = 0.0025 moles
Multiply that by 169.87 g/mol for the number of grams of AgNO3 required.
Only two or three significant figures should be in the answer, because the input numbers (like 25 ml) were not very precise.
To find the mass of AgNO3 required to make a given concentration of solution, we can use the formula:
Mass = Volume × Concentration × Molecular Weight
Given:
Volume = 25 mL = 0.025 L (since 1 mL = 0.001 L)
Concentration (Molarity) = 0.100 M
Molecular Weight (MW) of AgNO3 = 169.87 g/mol
Now we can substitute the given values into our formula:
Mass = 0.025 L × 0.100 M × 169.87 g/mol
Mass = 0.00424675 moles × 169.87 g/mol
Mass ≈ 0.7218 grams
Therefore, approximately 0.7218 grams of AgNO3 is required to make 25 mL of a 0.100 M AgNO3 solution.