An aqueous solution that is 30.0 percent sulfuric acid (H2SO4) by mass has a density of 1.105 g/mL. Determine the molarity of the solution.
I disagree with Damon's answer.
1.105 g/mL x 1000 mL x 0.30 x (1 mol/98g) = 3.4 M
I might make that 3.38 M to 3 s.f.
Well, it seems like we've got quite the chemical concoction here! Let's take a peek under the microscope and crunch some numbers.
To determine the molarity of the solution, we'll need a bit of a chemistry dance. First, we need to find the moles of sulfuric acid (H2SO4) in the solution. The molarity (M) is defined as the number of moles of solute per liter of solution.
Now, since we know the density of the solution is 1.105 g/mL, we can convert that to grams per liter. Multiplying the density by 1000 (to convert mL to L) gives us 1105 g/L.
Next, we calculate the mass of the sulfuric acid present in 1 liter of the solution. Since the solution is 30.0% sulfuric acid by mass, we multiply the mass of the solution by 0.3 to find the mass of the sulfuric acid. Therefore, the mass of the sulfuric acid is 1105 g/L * 0.3 = 331.5 g/L.
To find the moles of sulfuric acid, we divide the mass of sulfuric acid (in grams) by its molar mass. The molar mass of H2SO4 is (1.01 g/mol * 2) + 32.07 g/mol + (16.00 g/mol * 4) = 98.09 g/mol. Dividing 331.5 g/L by 98.09 g/mol gives us 3.38 mol/L.
So, the molarity of the sulfuric acid solution is approximately 3.38 M.
Now, I mustache you a question: Why was the math book sad? Because it had too many problems! I hope that answers your question and brings a smile to your face!
To determine the molarity of the solution, we need to know the concentration of solute (sulfuric acid) in moles per liter (mol/L).
First, we should find the mass of sulfuric acid in the solution. To do that, we assume we have 100 g of the solution (since percentage by mass is given). Therefore, the mass of sulfuric acid (H2SO4) in the solution will be 30.0 g (since it is 30.0% of 100 g).
Next, we need to convert the mass of sulfuric acid into moles. To do this, we divide the mass of sulfuric acid by its molar mass. The molar mass of sulfuric acid (H2SO4) can be calculated as follows:
H(2) + S + O(4) = (2 * 1.01 g/mol) + 32.07 g/mol + (4 * 16.00 g/mol) = 98.09 g/mol
So, 30.0 g H2SO4 is equal to (30.0 g / 98.09 g/mol) = 0.3059 mol H2SO4.
Finally, we can determine the volume of the solution in liters. The density of the solution is given as 1.105 g/mL. Therefore, for 100 g of solution, the volume will be (100 g / 1.105 g/mL) = 90.498 mL = 0.090498 L.
Now we have the moles of sulfuric acid (H2SO4) and the volume of the solution in liters, so we can calculate the molarity.
Molarity (M) = Moles of solute (H2SO4) / Volume of solution (L)
Molarity = 0.3059 mol / 0.090498 L = 3.38 M
Therefore, the molarity of the solution is 3.38 M.
water is 1 g/mL
so we have 0.105 g of H2SO4 per mL
How many mols is that
H2 = 2 g/mol
S = 32 g/mol
4 O = 64 g/mol
total = 98 g/mol
.105/98 = .0011 mol /mL which is
1.1 mol/Liter