HNO3 used as a reagent has a specific gravity of 1.42 g/ mL and contains 70% by strength HNO3. Calculate:
Volume of acid that contains 60 g pure acid?
Volume of water required to make 1 molar solution from 2 mL concentrated HNO3?
I have interpreted the problem differently from Bob Pursley so my answers differ.
First, the pure acid is what molarity?
1.42 x 1000 x 0.70 x (1/63) = 15.8M
Then use c1v1 = c2v2
15.8M x 2 mL = 1M x v2
v2 = about 15.8 x 2/1 = about 31.6 mL = total volume v2 which means you add about 31.6-2 = about 29.6 mL H2O
Then to obtain 60 g pure HNO3 you will want 1.42 g/mL x ?mL x 0.70 = 60 and solve for ? mL. I obtained something like 60g of the pure acid.
To solve these problems, we need to understand the relationship between concentration, volume, mass, and molarity.
1. Volume of acid that contains 60 g pure acid:
First, we need to determine the mass of the 70% HNO3 solution that contains 60 g of pure acid. Since the solution is 70% by strength, this means that for every 100 g of the solution, 70 g is pure acid.
Let's find the mass of the solution needed to have 60 g of pure acid:
Mass of solution = Mass of pure acid / (% strength / 100)
Mass of solution = 60 g / (70 / 100) = 85.7 g
Next, we can use the specific gravity to calculate the volume of the solution. Remember that the specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water). In this case, the specific gravity of the HNO3 solution is given as 1.42 g/mL.
Volume of solution = Mass of solution / Specific gravity
Volume of solution = 85.7 g / 1.42 g/mL = 60.28 mL
Therefore, to obtain 60 g of pure acid, you would need approximately 60.28 mL of the 70% HNO3 solution.
2. Volume of water required to make a 1 molar solution from 2 mL of concentrated HNO3:
To calculate the volume of water needed, we need to know the concentration of the concentrated HNO3 solution.
For the purpose of this explanation, let's assume the concentrated HNO3 solution is 100% pure HNO3.
To make a 1 molar (1 M) solution, the amount of solute (in this case, HNO3) needs to be equal to the molar concentration. Since 1 M is defined as 1 mole per liter of solution, we need to calculate the number of moles present in 2 mL of concentrated HNO3.
First, we convert the volume of HNO3 from milliliters to liters:
Volume of concentrated HNO3 = 2 mL = 2/1000 L = 0.002 L
Since we assumed the HNO3 is 100% pure, we can calculate the number of moles using the molar mass of HNO3 (63 g/mol).
Number of moles = Mass / Molar mass
Number of moles = 0.002 L * 63 g/mol = 0.126 mol
To make a 1 M solution, we need to dilute the HNO3 with water such that the final volume is 1 liter (1000 mL). By dilution, we keep the number of moles constant.
Now, let's calculate the volume of water needed:
Volume of water = Final volume - Volume of HNO3
Volume of water = 1000 mL - 2 mL = 998 mL
Therefore, to make a 1 M solution from 2 mL of concentrated HNO3, you would need approximately 998 mL of water.
60g/1.42g/ml= 42.25ml
Molarity concentrated sol= .060/63*1/42.25= 2.25
so you want to dilute it 2.25/1 times, which means one part concentrated HNO#, 1.25 part water.
One part is 2ml, so 1.25 part water is 2.50ml
60g/1.42g/ml= 42.25ml
Molarity concentrated sol= .060/63*1/42.25= 2.25
so you want to dilute it 2.25/1 times, which means one part concentrated HNO#, 1.25 part water.
One part is 2ml, so 1.25 part water is 2.50ml
First, the pure acid is what molarity?
1.42 x 1000 x 0.70 x (1/63) = 15.8M
Then use c1v1 = c2v2
15.8M x 2 mL = 1M x v2
v2 = about 15.8 x 2/1 = about 31.6 mL = total volume v2 which means you add about 31.6-2 = about 29.6 mL H2O
Then to obtain 60 g pure HNO3 you will want 1.42 g/mL x ?mL x 0.70 = 60 and solve for ? mL. I obtained something like 60g of the pure acid.