How many grams of HNO3 are required to prepare 500Ml of a 0.6601M HNO3 Solution?
To solve this problem, we can use the formula:
moles = concentration x volume
First, we need to calculate the number of moles of HNO3 required to prepare the solution:
moles = 0.6601 mol/L x 0.5 L = 0.33005 mol
Next, we can use the molar mass of HNO3 to convert moles to grams:
molar mass of HNO3 = 63.01 g/mol
grams = moles x molar mass = 0.33005 mol x 63.01 g/mol = 20.79 g
Therefore, 20.79 grams of HNO3 are required to prepare a 500 mL of 0.6601 M HNO3 solution.
To calculate the grams of HNO3 required to prepare the solution, we need to use the formula:
moles = concentration × volume
First, we need to convert the mL to liters:
500 mL = 500/1000 = 0.5 liters
Now, we can calculate the moles of HNO3 using the given concentration and volume:
moles = 0.6601 M × 0.5 L = 0.33005 moles
The molecular weight of HNO3 is approximately 63 g/mol. To find the grams of HNO3, we can multiply the moles by the molecular weight:
grams = moles × molecular weight
grams = 0.33005 moles × 63 g/mol ≈ 20.78 grams
Therefore, approximately 20.78 grams of HNO3 are required to prepare 500 mL of a 0.6601 M HNO3 solution.