The density of 19.0% by mass of nitric acid, HNO3 is 1.11 g cm-3. Calculate the
volume (dm3) of acid needed to prepare 2.00 dm3 of 3.00 mol dm-3 HNO3
solution. (FW HNO3 = 63 g/mol)
First determine the concn of HNO3 in the 19.0% stuff.
1.11 g/cc x 1000 cc x 0.19 x (1 mol/98 g) = estimated 2 mol/dm^3
Then use the dilution formula of
c1v1 = c2v2
2*v = 2*3
Solve for v = approx 3 dm^3
Remember to recalculate all of the above.
Well, I'm glad you didn't ask me to calculate how many clowns can fit into a car, because that's a whole different kind of chemistry! But hey, let's tackle this acid problem together.
First, we need to find the number of moles of HNO3 we need. The concentration of the solution is given as 3.00 mol dm-3, and the volume is 2.00 dm3. So, we can use the formula:
moles = concentration x volume
moles = 3.00 mol dm-3 x 2.00 dm3
moles = 6.00 mol
Alright, now we know we need 6.00 moles of HNO3. But the concentration of the nitric acid is given as 19.0% by mass. So, we need to find the mass of HNO3 first.
To find the mass, we can use the formula:
mass = concentration x volume x molar mass
mass = 0.190 x 2.00 dm3 x 63 g/mol
mass = 24.06 g
Ah, math! It certainly knows how to keep us on our toes. Now that we have the mass, we can calculate the volume of the acid needed. We know that the density of the acid is 1.11 g cm-3, which is equivalent to 1.11 g/mL.
volume = mass / density
volume = 24.06 g / 1.11 g/mL
volume = 21.7 mL
Oh, we're so close! But wait, we need to convert mL to dm3 since that's what the question is asking for. Remember that 1 dm3 is equal to 1000 mL.
volume = 21.7 mL / 1000
volume = 0.0217 dm3
And there you have it! The volume of nitric acid needed to prepare 2.00 dm3 of 3.00 mol dm-3 HNO3 solution is approximately 0.0217 dm3. Now that's an acid solution to the problem!
To calculate the volume of acid needed, we can use the formula:
Volume (dm^3) = (amount of solute / concentration)
First, let's calculate the amount of solute, which is the number of moles:
Amount of solute (mol) = concentration × volume
Given:
Concentration = 3.00 mol dm^-3
Volume = 2.00 dm^3
Amount of solute = 3.00 mol dm^-3 × 2.00 dm^3 = 6.00 mol
Next, let's convert the amount of solute to grams using the molecular weight:
Mass of solute (g) = Amount of solute (mol) × FW HNO3
Given:
FW HNO3 = 63 g/mol
Amount of solute = 6.00 mol
Mass of solute = 6.00 mol × 63 g/mol = 378 g
Now, let's calculate the density of nitric acid:
Density = mass / volume
Given:
Density = 1.11 g cm^-3
Since the density is given per cubic centimeter (cm^3), we need to convert the volume to cubic centimeters:
1 dm^3 = 1000 cm^3
Volume (cm^3) = 2.00 dm^3 × 1000 cm^3/dm^3 = 2000 cm^3
Now we can calculate the mass of the 2.00 dm^3 of HNO3 solution:
Mass = density × volume
Mass = 1.11 g cm^-3 × 2000 cm^3 = 2220 g
Finally, let's calculate the volume (in dm^3) of the acid needed to prepare the solution:
Volume (dm^3) = mass of solute / density
Volume = 378 g / 1.11 g cm^-3 = 340.54 cm^3
Since 1 dm^3 = 1000 cm^3, the volume in dm^3 is:
Volume (dm^3) = 340.54 cm^3 / 1000 cm^3/dm^3 = 0.341 dm^3
Therefore, the volume of acid needed to prepare 2.00 dm^3 of 3.00 mol dm^-3 HNO3 solution is approximately 0.341 dm^3.
To solve this problem, we need to first calculate the mass of the nitric acid needed to prepare a 2.00 dm3 solution of 3.00 mol dm-3 HNO3. Then we can use the density of the nitric acid to determine the volume required.
First, let's calculate the mass of HNO3 needed:
The molar mass of HNO3 is 63 g/mol.
The concentration of the solution is 3.00 mol dm-3.
To calculate the mass, we can use the formula:
mass = molar mass x volume of solution
mass = 63 g/mol x 2.00 dm3 = 126 g
Now that we have the mass, we can use the density of the nitric acid to calculate the volume needed.
The density is given as 1.11 g cm-3, which is equivalent to 1.11 g/mL since 1 mL is equal to 1 cm3.
To calculate the volume, we can rearrange the density formula:
volume = mass / density
volume = 126 g / 1.11 g/mL = 113.51 mL
Finally, we convert the volume from mL to dm3:
volume in dm3 = 113.51 mL x (1 dm3 / 1000 mL) = 0.11351 dm3
Therefore, the volume of nitric acid needed to prepare a 2.00 dm3 solution of 3.00 mol dm-3 HNO3 is approximately 0.11351 dm3.