Concentrated sulfuric acid is 18.4 molar and has a density of 1.84 grams per milliliter. After dilution with water to 5.20 molar the solution has a density of 1.38 grams per milliliter. This concentration of H2SO4 is often used in lead storage batteries(like a car battery).
1.Calculate the volume of concentrated H2SO4 needed to make 1.00L of 5.20 molar solution.
2.Calculate the % concentration H2SO4 in the concentrated solution.
3.Calculate the volume of 5.20 molar H2SO4 required to react completely with 100.0 grams of NaHCO3. (Carbon dioxide and water are two of the three products of this reaction-can you figure out the other?)
4.What is the molality of the 5.20 molar H2SO4 solution?
I understand 3 and 4.
1 I don't understand mL x M = mL x M ??
2. Is number 2. (1803.2/1840)x100=98%H2SO4
mL x M = mL x M is the dilution formula.
mL of the concd acid is the unknown.
M of the concd acid is 18.4 M
mL of the diluted solution is 1,000 mL (you want 1.00 L). M of the diluted solution is 5.20M. Solve for the only unknown which is mL of the concd acid you need to use to dilute to 1.00 L to make the 5.20L..
Yes, it is 98% H2SO4.
This post is so far down that any follow up questions should be directed at the top of the page (make a new post that is) so someone will see it sooner.
Let's break down each question and explain how to solve them:
1. To calculate the volume of concentrated H2SO4 needed to make 1.00L of 5.20 molar solution, we’ll use the equation mL x M = mL x M, where mL represents the volume of the concentrated H2SO4 and M represents the molarity. In this case, we have concentrated H2SO4 with a molarity of 18.4 M and we want to make a 5.20 M solution. The volume of the concentrated H2SO4 can be calculated as:
(18.4 M) x (V1 mL) = (5.20 M) x (1000 mL)
Solving for V1, we get:
V1 mL = (5.20 M) x (1000 mL) / (18.4 M)
V1 mL ≈ 282.61 mL
Therefore, you would need approximately 282.61 mL of concentrated H2SO4 to make 1.00L of a 5.20 M solution.
2. To calculate the % concentration of H2SO4 in the concentrated solution, we need to determine the concentration in grams per 100 mL. Since the density of the concentrated H2SO4 is given as 1.84 g/mL, we'll use the equation:
% concentration = (mass of solute / volume of solution) x 100
Given that the density of the concentrated H2SO4 is 1.84 g/mL, we can calculate the mass of 100 mL of the solution as:
mass = density x volume
mass = 1.84 g/mL x 100 mL
mass = 184 g
Now, to find the % concentration, we'll use the equation:
% concentration = (mass of H2SO4 / mass of solution) x 100
Since the molar mass of H2SO4 is 98 g/mol, the mass of H2SO4 in the concentrated solution is:
mass of H2SO4 = (18.4 M) x (98 g/mol) = 1803.2 g
% concentration = (1803.2 g / 184 g) x 100
% concentration ≈ 980.87%
Therefore, the % concentration of H2SO4 in the concentrated solution is approximately 980.87%.
3. This question involves the stoichiometry of a reaction between H2SO4 and NaHCO3. When H2SO4 reacts with NaHCO3, it forms CO2 (carbon dioxide), H2O (water), and Na2SO4 (sodium sulfate).
The balanced equation for the reaction is:
H2SO4 + 2 NaHCO3 -> CO2 + H2O + Na2SO4
To calculate the volume of 5.20 M H2SO4 required to react completely with 100.0 grams of NaHCO3, we need to convert the mass of NaHCO3 to moles and then use stoichiometry to determine the molar ratio between H2SO4 and NaHCO3.
First, calculate the number of moles of NaHCO3:
moles of NaHCO3 = (mass of NaHCO3 / molar mass of NaHCO3)
moles of NaHCO3 = (100.0 g / 84.01 g/mol) ≈ 1.19 mol
According to the balanced equation, 1 mole of H2SO4 reacts with 2 moles of NaHCO3. Therefore, to completely react with 1.19 moles of NaHCO3, we will need:
moles of H2SO4 = (1.19 mol) x (1 mol H2SO4 / 2 mol NaHCO3)
moles of H2SO4 ≈ 0.595 mol
Now, we can calculate the volume of 5.20 M H2SO4 using the equation mL x M = moles x 1000:
(5.20 M) x (V2 mL) = (0.595 mol) x (1000 mL)
V2 mL ≈ 114.42 mL
Therefore, approximately 114.42 mL of 5.20 M H2SO4 is required to react completely with 100.0 grams of NaHCO3.
4. The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent (in kg). In this case, the solute is H2SO4 and the solvent is water.
To find the molality of the 5.20 M H2SO4 solution, we need to calculate the moles of H2SO4 and the mass of water. The moles of H2SO4 can be calculated using the molarity and volume of the solution:
moles of H2SO4 = (5.20 M) x (1.00 L)
moles of H2SO4 = 5.20 mol
The density of the diluted solution is given as 1.38 g/mL, which is equivalent to 1.38 kg/L. Therefore, the mass of water in 1.00 L of solution is:
mass of water = (1.38 kg/L) x (1.00 L)
mass of water = 1.38 kg
Now we can calculate the molality using the equation:
molality = (moles of solute / mass of solvent in kg)
molality = (5.20 mol) / (1.38 kg)
molality ≈ 3.768 mol/kg
Therefore, the molality of the 5.20 M H2SO4 solution is approximately 3.768 mol/kg.