How many milliliters of concentrated nitric acid, HNO3, 70 % (wt/wt) are required to prepare 1 liter of 0.250 M solution? The molar mass and density of HNO3 are 63.0 g/mol and 1.36 g/cm respectively.
First, what is the M of the 70% HNO3. That's
1.36 g/mL x 1000 mL x 0.70 x (1/63) = ? M1
Then M1 x mL1 = M2 x mL2
M2 is 0.250; mL2 = 1000 mL, solve for mL1.
First, what is the M of the 70% HNO3. That's
1.36 g/mL x 1000 mL x 0.70 x (1/63) = ? M1
Then M1 x mL1 = M2 x mL2
M2 is 0.250; mL2 = 1000 mL, solve for mL1
To solve this problem, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molarity of the concentrated acid
V1 = initial volume of the concentrated acid
M2 = final molarity of the diluted acid
V2 = final volume of the diluted acid
First, let's calculate the number of moles of nitric acid needed to prepare the 1 liter of 0.250 M solution:
Molarity (M) = moles/volume (L)
0.250 M = moles/1 L (Equation 1)
Rearranging Equation 1, we can solve for moles:
moles = 0.250 M x 1 L
moles = 0.250 mol
Now, let's calculate the volume of the concentrated nitric acid needed. We'll use the equation:
moles = mass/molar mass
Rearranging this equation, we get:
mass = moles x molar mass
mass = 0.250 mol x 63.0 g/mol
mass = 15.75 g
Next, let's calculate the volume of the concentrated nitric acid needed. We'll use the equation:
density = mass/volume
Rearranging this equation, we get:
volume = mass/density
volume = 15.75 g / 1.36 g/cm³
volume = 11.57 cm³
However, we need the volume in milliliters. Since 1 cm³ = 1 mL, the volume is also 11.57 mL.
Therefore, you would need to use 11.57 milliliters of concentrated nitric acid to prepare 1 liter of 0.250 M solution.