Concentrated sulfuric acid (18.4 molar H2SO4) 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 and can be used as an electrolyte in lead storage batteries for automobiles.

1. Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20 molar H2SO4.

2. Determine the mass percent of H2SO4 in the original concentrated solution.

3. Calculate the volume of 5.20 molar H2SO4 that can be completely neutralized with 10.5 grams of sodium bicarbonate, NaHCO3.

4. What is the molality of the 5.20 molar H2SO4.

1) how many moles is in 1Liter of 5.2M? What volume of 18.5M will have that many moles?

2) You know the mass from density. The mass consists of water and acid. You know how much water is there

1.84g/ml= 1g/ml water + .84gacid/ml

3) Use the titration equation.
4) used density to figure the mass of the solvent.

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To answer these questions, we need to use some calculations based on the given information. Let's break down each question and explain the steps to solve them.

1. Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20 molar H2SO4.

To calculate the volume of concentrated acid required, we need to use the dilution formula:

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

We know that the final volume of the diluted acid is 1.00 liter (V2) and the final molarity is 5.20 molar (M2). We are looking for the initial volume of the concentrated acid (V1). The initial molarity of the concentrated acid (M1) is given as 18.4 molar.

18.4 M * V1 = 5.20 M * 1.00 L

Simplifying the equation:

V1 = (5.20 M * 1.00 L) / 18.4 M

So, the volume of concentrated acid required to prepare 1.00 liter of 5.20 molar H2SO4 is approximately 0.283 liters.

2. Determine the mass percent of H2SO4 in the original concentrated solution.

To calculate the mass percent of H2SO4, we need to know the mass of H2SO4 and the mass of the solution. We can calculate the mass of H2SO4 using the molar mass of H2SO4 (98.09 g/mol) and the molarity (18.4 M) of the concentrated acid.

First, we calculate the mass of H2SO4 in one liter of the concentrated acid solution:

mass_H2SO4 = molarity * molar_mass * volume_solution

mass_H2SO4 = 18.4 M * 98.09 g/mol * 1.00 L

Next, we calculate the mass of the solution using the density of the concentrated acid (1.84 g/mL) and the volume of the solution:

mass_solution = density * volume_solution

mass_solution = 1.84 g/mL * 1.00 L

Finally, we can calculate the mass percent of H2SO4:

mass_percent_H2SO4 = (mass_H2SO4 / mass_solution) * 100

Plugging in the values:

mass_percent_H2SO4 = (18.4 M * 98.09 g/mol * 1.00 L) / (1.84 g/mL * 1.00 L) * 100

The mass percent of H2SO4 in the original concentrated solution is approximately 99.974%.

3. Calculate the volume of 5.20 molar H2SO4 that can be completely neutralized with 10.5 grams of sodium bicarbonate, NaHCO3.

To calculate the volume of H2SO4 that can be neutralized, we need to use the stoichiometry between H2SO4 and NaHCO3. The balanced equation for the reaction is:

H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + 2CO2

From the chemical equation, we can see that the molar ratio between H2SO4 and NaHCO3 is 1:2.

First, we calculate the number of moles of NaHCO3:

moles_NaHCO3 = mass_NaHCO3 / molar_mass_NaHCO3

moles_NaHCO3 = 10.5 g / 84.01 g/mol

Since the molar ratio between H2SO4 and NaHCO3 is 1:2, the number of moles of H2SO4 needed is equal to half the number of moles of NaHCO3:

moles_H2SO4 = moles_NaHCO3 / 2

Next, we can calculate the volume of 5.20 M H2SO4 that can react with the calculated moles of H2SO4.

volume_H2SO4 = moles_H2SO4 / molarity_H2SO4

volume_H2SO4 = moles_H2SO4 / 5.20 M

Plugging in the values:

volume_H2SO4 = (10.5 g / 84.01 g/mol) / 5.20 M

The volume of 5.20 M H2SO4 that can be completely neutralized with 10.5 grams of NaHCO3 is approximately 0.12 liters.

4. What is the molality of the 5.20 molar H2SO4.

Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is H2SO4, and the solvent is water.

To calculate the molality, we need to convert the molarity to moles of solute and then divide by the mass of the solvent in kilograms.

First, we calculate the moles of H2SO4:

moles_H2SO4 = molarity_H2SO4 * volume_solution

moles_H2SO4 = 5.20 M * 1.00 L

Next, we need to convert the mass of water to kilograms. The density of water is 1.00 g/mL.

mass_water = density_water * volume_solution

mass_water = 1.00 g/mL * 1.00 L

Now we can calculate the molality:

molality_H2SO4 = moles_H2SO4 / mass_water (in kg)

molality_H2SO4 = moles_H2SO4 / (mass_water / 1000)

Plugging in the values:

molality_H2SO4 = (5.20 M * 1.00 L) / ((1.00 g/mL * 1.00 L) / 1000)

The molality of the 5.20 molar H2SO4 solution is approximately 5.20 mol/kg.