Consider 18.0 M H2SO4, if you need to make 250.0 mL of a 3.0 M solution of H2SO4, how would you do this?
18.0 M H2SO4 has a density of 1.84g/mL what is the molality of solution? Mass % of H2SO4 in the solution and the mol fraction of H2SO4 and water in the solution?
I went back and looked at my response to this last night. Since you have reposted the entire problem I don't know what you understood and what you did not understand. It's non-sense for me to redo the whole thing if you need just a part of it. Explain in detail your problem(s) with this.
DrBob I wrote back and wanted to know if I did it right can you check my answers please
done
You won't really know how much it helped until you try one on your own.
To make a 3.0 M solution of H2SO4 with 18.0 M H2SO4, you will need to dilute the 18.0 M solution to the desired concentration. Here's how you can do this:
1. Start by calculating the volume of the concentrated acid solution that you will need to make the desired solution. This can be done using the dilution equation:
M1V1 = M2V2
Where:
M1 = Initial concentration of the concentrated acid solution (18.0 M)
V1 = Volume of the concentrated acid solution (unknown)
M2 = Final concentration of the diluted acid solution (3.0 M)
V2 = Volume of the diluted acid solution (250.0 mL or 0.250 L)
Rearranging the equation to solve for V1:
V1 = (M2 × V2) / M1
V1 = (3.0 M × 0.250 L) / 18.0 M
V1 ≈ 0.0417 L or 41.7 mL
So, you will need approximately 41.7 mL of the 18.0 M H2SO4 solution.
2. Measure 41.7 mL of the 18.0 M H2SO4 solution using a graduated cylinder or a pipette.
3. Transfer the measured volume of the 18.0 M H2SO4 solution to a volumetric flask.
4. Add distilled water to the volumetric flask until the total volume reaches 250.0 mL. Make sure to mix the solution thoroughly to ensure uniformity.
Now, let's move on to the other questions regarding the properties of the solution:
1. To calculate the molality of the solution, we need to know the mass of the solute (H2SO4) and the mass of the solvent (water).
- The molar mass of H2SO4 is:
2(1.01 g/mol of H) + 32.07 g/mol of S + 4(16.00 g/mol of O) = 98.09 g/mol
- The density of the 18.0 M H2SO4 is given as 1.84 g/mL. Therefore, we can calculate the mass of H2SO4 in the solution by multiplying the density by the volume of the solution (41.7 mL):
Mass of H2SO4 = 1.84 g/mL × 41.7 mL ≈ 76.748 g
- Now, we can calculate the molality using the formula:
Molality = (moles of solute) / (mass of solvent in kg)
Since the solution is prepared in 250.0 mL (0.250 L), the mass of water can be calculated:
Mass of water = density of water × volume of water
= 1 g/mL × (250 - 41.7) mL
= 226.3 g
Converting the mass of water to kg:
Mass of water = 226.3 g / 1000 g/kg
= 0.2263 kg
Now, we can calculate the molality:
Molality = (mol of H2SO4) / (mass of water in kg)
= (76.748 g / 98.09 g/mol) / 0.2263 kg
≈ 3.46 mol/kg
The molality of the solution is approximately 3.46 mol/kg.
2. To find the mass percent of H2SO4 in the solution, we need to divide the mass of H2SO4 by the total mass of the solution and multiply by 100:
Mass percent of H2SO4 = (mass of H2SO4 / total mass of solution) × 100
The total mass of the solution can be calculated by summing the mass of H2SO4 and the mass of water:
Total mass of solution = mass of H2SO4 + mass of water
= 76.748 g + 226.3 g
= 303.048 g
Plugging these values into the formula:
Mass percent of H2SO4 = (76.748 g / 303.048 g) × 100
≈ 25.34%
Therefore, the solution contains approximately 25.34% H2SO4 by mass.
3. The mole fraction of H2SO4 and water can be calculated using their respective moles and the total moles in the solution.
The moles of H2SO4 can be calculated using the formula:
Moles of H2SO4 = (mass of H2SO4) / (molar mass of H2SO4)
Moles of H2SO4 = 76.748 g / 98.09 g/mol ≈ 0.782 mol
The moles of water can be found by dividing its mass by its molar mass:
Moles of water = (mass of water) / (molar mass of water)
Moles of water = 226.3 g / 18.015 g/mol ≈ 12.563 mol
The total moles in the solution are the sum of the moles of H2SO4 and water:
Total moles = moles of H2SO4 + moles of water
= 0.782 mol + 12.563 mol
≈ 13.345 mol
Finally, we can calculate the mole fraction:
Mole fraction of H2SO4 = (moles of H2SO4) / (total moles)
= 0.782 mol / 13.345 mol
≈ 0.0586
Mole fraction of water = (moles of water) / (total moles)
= 12.563 mol / 13.345 mol
≈ 0.9414
Therefore, the mole fraction of H2SO4 in the solution is approximately 0.0586, and the mole fraction of water is approximately 0.9414.