a piece of Aluminium weighing 2.7g is titrated with 75 mL of H2SO4(specific gravity 1.18 and 24.7% H2SO4 by weight). after the metal is completely dissolved the solution is diluted to 400 mL . calculate the molarity of free H2SO4 in the solution.

2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2

mols H2SO4 initially = 1.18 g/mL x 75 mL x (24.7/100) x 1/98 = approx 0.228
mols Al initially = 2.7g/27 = 0.1 mol

All of the Al is dissolved. How much of the H2SO4 is used? That is
0.1 mol Al x (3 mols H2SO4/2 mols Al) = 0.15 mols H2SO4 used.
How much H2SO4 remains unreacted? That's 0.228 - 0.150 = ?
M H2SO4 in final solution is mols/L = mols/0.400 L = ?
Post your work if you get stuck.

I got the solution from you. I was stucked for hours. You saved my day. A heartful thanks to you!!! Keep growing.

To calculate the molarity of free H2SO4 in the solution, we need to determine the number of moles of H2SO4 present in the solution.

1. Calculate the number of moles of H2SO4 in the 75 ml of H2SO4 solution:
- We know that the specific gravity of H2SO4 is 1.18 and the weight percentage of H2SO4 is 24.7%.
- The weight percentage can be converted to weight by assuming 100g of the solution. Therefore, there would be 24.7g of H2SO4.
- The molar mass of H2SO4 is 98.09 g/mol.
- Mass of H2SO4 in the 75 ml of solution can be calculated as: (24.7 g/100 ml) x 75 ml = 18.525 g.
- The number of moles of H2SO4 in the 75 ml of solution can be calculated as: (18.525 g / 98.09 g/mol) = 0.1887 mol.

2. Calculate the volume of the final diluted solution containing the 0.1887 mol of H2SO4:
- The solution was diluted to 400 ml.

3. Calculate the molarity (M) of H2SO4 in the solution:
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- The volume of the solution after dilution is 400 ml, which is equivalent to 0.4 L.
- Therefore, the molarity (M) of H2SO4 in the solution is: (0.1887 mol / 0.4 L) = 0.4718 M.

Therefore, the molarity of free H2SO4 in the solution is 0.4718 M.

To calculate the molarity of free H2SO4 in the solution, we first need to determine the number of moles of H2SO4 present in the solution. Here's how you can do it:

Step 1: Calculate the number of moles of Aluminium (Al).

To calculate the moles of Aluminium, use the formula:

Number of moles = mass / molar mass

Given that the mass of Aluminium is 2.7g, and the molar mass of Aluminium is 26.98 g/mol, we can calculate the moles of Aluminium:

Number of moles of Al = 2.7g / 26.98 g/mol = 0.100 moles of Al

Step 2: Calculate the number of moles of H2SO4.

Since we know the volume and concentration of H2SO4 in mL and the specific gravity, we need to convert mL to grams:

Number of moles of H2SO4 = (volume in mL) * (specific gravity) * (percentage of H2SO4 by weight) / (molar mass)

Given that the volume of H2SO4 is 75 mL, specific gravity is 1.18, the percentage of H2SO4 by weight is 24.7%, and the molar mass of H2SO4 is 98.09 g/mol, we can calculate the moles of H2SO4:

Number of moles of H2SO4 = (75 mL) * (1.18) * (0.247) / (98.09 g/mol) = 0.182 moles of H2SO4

Step 3: Calculate the total number of moles of solute in the solution.

To calculate the total number of moles of solute in the solution, add the moles of Al and moles of H2SO4:

Total moles of solute = moles of Al + moles of H2SO4
= 0.100 moles + 0.182 moles
= 0.282 moles

Step 4: Calculate the molarity of H2SO4.

Molarity (M) is defined as moles of solute divided by the volume of solution in liters. Since the solution is diluted to 400 mL (which is 0.4 L), we can calculate the molarity of H2SO4:

Molarity of H2SO4 = total moles of solute / volume of solution in liters
= 0.282 moles / 0.4 L
= 0.705 M

Therefore, the molarity of free H2SO4 in the solution is 0.705 M.