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?

1.

mL x M = mL x M

2. How much does 1 L of the solution weigh.
1.84*1000 mL = 1840
How many grams H2SO4 are there? 18.4 moles x 98 g/mol = 1803.2 (confirm that).
%H2SO4 = (g H2SO4/g solution)*100 = ??
3. This is a stoichiometry problem. Here is a solved example problem. Just follow the steps. (The other product is Na2SO4.)
http://www.jiskha.com/science/chemistry/stoichiometry.html
4. Take 1,000 mL of the solution. That's 5.2 M = 5.2 moles H2SO4/L soln
How many g H2SO4 are in the soln? That's 5.2 x 98 g H2SO4/mol = 509.6 g
How much does the soln weigh? That's 1.38 g/mL x 1000 mL = 1380 g.
How much does the water weigh? That's 1380 g - 509.6 = ??
molality = 5.2 moles/L soln.
My answer is about 5.97 but you should confirm that.

370 mL

1. To calculate the volume of concentrated H2SO4 needed to make 1.00L of 5.20 molar solution, we can use the formula:

M1V1 = M2V2

Where:
M1 = initial molarity of the concentrated acid
V1 = volume of the concentrated acid
M2 = final molarity of the diluted solution
V2 = final volume of the diluted solution

Plugging in the values:
M1 = 18.4 M
V1 = ?
M2 = 5.20 M
V2 = 1.00 L = 1000 mL

Using the formula, we get:
18.4 M * V1 = 5.20 M * 1000 mL

Solving for V1:
V1 = (5.20 M * 1000 mL) / 18.4 M ≈ 282.61 mL

Therefore, approximately 282.61 mL of concentrated H2SO4 is needed to make 1.00L of 5.20 molar solution.

2. The % concentration of H2SO4 in the concentrated solution can be calculated using the formula:

% concentration = (mass of solute / mass of solution) * 100%

Since we know the density of the concentrated sulfuric acid (1.84 g/mL), we can find the mass of the concentrated solution in 1L:

mass of solution = density * volume = 1.84 g/mL * 1000 mL = 1840 g

To find the mass of H2SO4 in the solution, we can use the molarity (18.4 M) and the molar mass of H2SO4 (98.09 g/mol):

mass of H2SO4 = molarity * volume * molar mass = 18.4 M * 1000 mL * 98.09 g/mol

Mass of H2SO4 = 1.80 x 10^6 g

Now we can calculate the % concentration:

% concentration = (mass of H2SO4 / mass of solution) * 100% = (1.80 x 10^6 g / 1840 g) * 100% ≈ 97.83%

Therefore, the % concentration of H2SO4 in the concentrated solution is approximately 97.83%.

3. To calculate the volume of 5.20 molar H2SO4 required to react completely with 100.0 grams of NaHCO3, we first need to write the balanced chemical equation for the reaction:

2 NaHCO3 + H2SO4 -> Na2SO4 + 2CO2 + 2H2O

From the equation, we can see that one mole of H2SO4 reacts with 2 moles of NaHCO3.

First, let's calculate the number of moles of NaHCO3:

moles of NaHCO3 = mass of NaHCO3 / molar mass of NaHCO3

The molar mass of NaHCO3 is:
23.0 g/mol (Na) + 1.0 g/mol (H) + 12.0 g/mol (C) + 48.0 g/mol (O) + 16.0 g/mol (O) = 84.0 g/mol

moles of NaHCO3 = 100.0 g / 84.0 g/mol ≈ 1.19 mol

Since the reaction ratio is 2:1 (NaHCO3 to H2SO4), we need twice as many moles of H2SO4.

moles of H2SO4 = 2 * moles of NaHCO3 ≈ 2.38 mol

Finally, we can calculate the volume of 5.20 molar H2SO4 required using the formula:

volume = moles / molarity

volume = 2.38 mol / 5.20 mol/L ≈ 0.46 L = 460 mL

Therefore, approximately 460 mL of 5.20 molar H2SO4 is required to react completely with 100.0 grams of NaHCO3.

The other product of this reaction is Na2SO4.

4. The molality (m) of a solution can be calculated using the formula:

molality = moles of solute / mass of solvent (in kg)

In this case, the solute is H2SO4, and the solvent is water.

The molarity of the solution is given as 5.20 M, which means there are 5.20 moles of H2SO4 per liter of solution.

Since 1L of water weighs approximately 1000g or 1kg, we can calculate the molality:

moles of H2SO4 = molarity * volume = 5.20 M * 1 L = 5.20 mol
mass of solvent = 1 kg

molality = 5.20 mol / 1 kg ≈ 5.20 m

Therefore, the molality of the 5.20 molar H2SO4 solution is approximately 5.20 m.

1. To calculate the volume of concentrated H2SO4 needed to make 1.00L of 5.20 molar solution, you can use the formula:

M1V1 = M2V2

Where:
M1 = initial molarity of the concentrated H2SO4
V1 = volume of the concentrated H2SO4 you want to find
M2 = final molarity of the diluted solution (5.20 M)
V2 = final volume of the diluted solution (1.00 L)

Rearranging the formula, we have:

V1 = (M2 * V2) / M1

Substituting the values,

V1 = (5.20 M * 1.00 L) / 18.4 M = 0.283 L

Therefore, you would need 0.283 L (or 283 mL) of the concentrated H2SO4 to make 1.00 L of the 5.20 molar solution.

2. To calculate the % concentration of H2SO4 in the concentrated solution, you can use the formula:

% concentration = (mass of solute / total mass of solution) * 100

The total mass of the solution is equal to the density (1.84 g/mL) multiplied by the volume of the solution.

Total mass = density * volume = 1.84 g/mL * 1.00 L = 1840 g

The mass of the solute (H2SO4) can be calculated using its molarity and molar mass (98.09 g/mol):

Mass of solute = molarity * volume * molar mass = 18.4 M * 1.00 L * 98.09 g/mol = 1802 g

Using the formula, we have:

% concentration = (1802 g / 1840 g) * 100 = 97.93%

Therefore, the concentrated H2SO4 solution has a % concentration of approximately 97.93%.

3. To calculate the volume of 5.20 molar H2SO4 required to react completely with 100.0 grams of NaHCO3, you need to determine the stoichiometry of the reaction between H2SO4 and NaHCO3.

The balanced chemical equation for the reaction is:

H2SO4 + 2NaHCO3 -> Na2SO4 + 2CO2 + 2H2O

From the equation, we can see that 2 moles of NaHCO3 react with 1 mole of H2SO4.

First, convert the mass of NaHCO3 to moles using its molar mass:

Moles of NaHCO3 = Mass / molar mass = 100.0 g / 84.01 g/mol = 1.19 mol

Since the molar ratio between H2SO4 and NaHCO3 is 1:2, we need twice the amount of moles of NaHCO3 in H2SO4:

Moles of H2SO4 = 2 * Moles of NaHCO3 = 2 * 1.19 mol = 2.38 mol

Now, we can use the formula:

Molarity = moles / volume

To find the volume of 5.20 M H2SO4, rearrange the formula:

Volume = moles / molarity = 2.38 mol / 5.20 M = 0.458 L or 458 mL

Therefore, 458 mL of 5.20 M H2SO4 is required to react completely with 100.0 grams of NaHCO3.

The third product of the reaction is Na2SO4 (sodium sulfate).

4. The molality of a solution is defined as the number of moles of solute per kilogram of solvent. Since water is the solvent and its density is known, we can calculate the molality of the 5.20 M H2SO4 solution.

First, calculate the mass of water needed for 1.00 L of the solution:

Mass of water = density of water * volume of water = 1.00 g/mL * 1000 mL = 1000 g

Next, convert the mass of water to kilograms:

Mass of water in kg = 1000 g / 1000 = 1.00 kg

Finally, calculate the molality using the formula:

Molality = moles of solute / mass of solvent (in kg)

To find the moles of H2SO4, use the formula:

Moles of H2SO4 = Molarity * Volume = 5.20 M * 1.00 L = 5.20 moles

Therefore, the molality is:

Molality = 5.20 moles / 1.00 kg = 5.20 mol/kg

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