A commercial sample of concentrated aqueous nitric acid (HNO3) is 18.2% nitric acid base mass and has a density of 1.12 g/mL.
(a) What is the molarity (M) of the HNO3 solution?
(b) What is the molality (m) of the HNO3 solution?
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I know that....
(a) molarity = moles of solute/liters of solution
(b) molality = moles of solute/kg of solvent
But, I don't know how to find the moles of HNO3 (63 amu) with the information I have been given.
assume some volume of the acid (it wont matter, it divides out).
Assume a volume of 2liters.
massnitric=.182*density*volume and you know density, and volume.
Now compute the moles of nitric given mass and its mole mass.
So then the mass would equal .407 g?
And that would be .401/63 = .0065 mol?
Is that correct so far?
I don't really understand how I am able to just pic a random value for volume. The volume effects the mass, which effects the moles, right?
You are not correct so far. I think you must have multiplied 2 L x 0.182 x 1.12. Note that the density is in g/mL so the units don't come out right.
To find the moles of HNO3, you need to use the given information about the concentration and density of the solution. Here's how you can calculate it step by step:
(a) To find the molarity (M) of the HNO3 solution:
1. Convert the given concentration of HNO3 from mass percent to grams per milliliter:
18.2% nitric acid by mass means that there are 18.2 grams of HNO3 in 100 grams of the solution.
This can be written as 18.2 g HNO3 / 100 g solution.
2. Use the density of the solution to convert grams to milliliters:
Given that the density is 1.12 g/mL, divide the total mass by the density:
18.2 g HNO3 / (1.12 g/mL) = 16.25 mL HNO3
3. Convert milliliters to liters:
16.25 mL / 1000 mL/L = 0.01625 L HNO3
4. Calculate the moles of HNO3:
To do this, you need to know the molar mass of HNO3, which is 63 g/mol.
Moles of HNO3 = mass of HNO3 (in grams) / molar mass of HNO3
= 18.2 g / 63 g/mol
5. Finally, calculate the molarity using the formula:
Molarity (M) = Moles of solute / Liters of solution
= (18.2 g / 63 g/mol) / 0.01625 L
(b) To find the molality (m) of the HNO3 solution:
1. The molality (m) is defined as moles of solute per kilogram of solvent. In this case, the solute is HNO3 and the solvent is water.
2. To find the moles of HNO3, you can use the same value calculated in part (a) (moles of HNO3 = 18.2 g / 63 g/mol).
3. To calculate the mass of water in kilograms, you need to convert the volume (in milliliters) of the solution:
Volume of solution = 16.25 mL.
The density of the solution is 1.12 g/mL, so:
Mass of solution = density x volume
= 1.12 g/mL x 16.25 mL
Since you're looking for the mass of water, subtract the mass of HNO3:
Mass of water = Mass of solution - Mass of HNO3
4. Convert the mass of water from grams to kilograms:
Mass of water (in kg) = Mass of water (in g) / 1000 g/kg
5. Finally, calculate the molality using the formula:
Molality (m) = Moles of solute / Mass of water (in kg)
= (18.2 g / 63 g/mol) / (Mass of water (in g) / 1000 g/kg)
By following these steps, you can find the molarity (M) and molality (m) of the HNO3 solution.