how much of 55% nitric acid, specific gravity 1g/ml is needed to prepare 100ml of 0.01M Nitric acid
First, calculate the molarity of the HNO3 solution you have.
That's 1000 mL x 1 g/mL x 0.55 x (1 mol/63 g) = 8.73 M
Then use the dilution formula of
mL1 x M1 = mL2 x M2
mL1 x 8.73 = 100 x 0.01
Solve for mL1, transfer that amount of the HNO3 to a 100 mL volumetric flask, add distilled water to the mark on the flask, stopper, mix thoroughly, label. done.
To prepare 100ml of 0.01M Nitric acid from 55% nitric acid with a specific gravity of 1g/ml, we need to calculate the required volume of the 55% nitric acid.
Step 1: Calculate the number of moles of nitric acid required.
Molarity = moles/volume
0.01M = moles/0.1L (since we need 100ml of 0.01M solution)
moles = 0.01M * 0.1L = 0.001 moles
Step 2: Determine the concentration of the 55% nitric acid.
The 55% nitric acid means it contains 55g of nitric acid in 100ml of solution. Since the specific gravity is 1g/ml, we can assume 100ml of the solution weighs 100g. Therefore, the 55% nitric acid has 55g of nitric acid.
Step 3: Convert the grams of nitric acid to moles.
Molar mass of nitric acid (HNO3) = 1.01g + 14.01g + 3(16.00g) = 63.01g/mol
moles = 55g / 63.01g/mol = 0.873 moles
Step 4: Determine the volume of the 55% nitric acid required.
Volume = moles/concentration
Volume = 0.001 moles / 0.873 moles/ml = 0.00114 L = 1.14 ml
Therefore, you would need approximately 1.14 ml of 55% nitric acid to prepare 100ml of 0.01M Nitric acid.
To calculate the amount of 55% nitric acid needed to prepare 100ml of 0.01M nitric acid, we need to consider the molarity and the specific gravity of the 55% nitric acid.
First, let's determine the concentration of the 55% nitric acid solution. A 55% nitric acid solution means that 55g of nitric acid is present in 100ml of solution. Since the specific gravity is given as 1g/ml, the solution's density will be 1g/ml.
Next, we need to convert the concentration of nitric acid to moles per liter (Molarity). The molar mass of nitric acid (HNO3) is approximately 63g/mol. Since the concentration is 0.01M (moles per liter), we have:
0.01 mol/L = (mass of nitric acid in solution) / (volume of solution in liters)
To find the mass of nitric acid required, we use the equation:
mass = concentration (in moles per liter) * molar mass * volume (in liters)
Given that we want to prepare 100ml of 0.01M nitric acid, which is 0.1 liters, we can calculate the mass of nitric acid as follows:
mass = 0.01 mol/L * 63 g/mol * 0.1 L
mass = 0.063 g
Therefore, we would need approximately 0.063 grams of 55% nitric acid, specific gravity 1g/ml, to prepare 100ml of 0.01M nitric acid.