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?
#1.
mL1 x M1 = mL2 x M2
Substitute and solve.
250mL x 3.0M = mL2 x 18
#2.
18.0M means 18.0 mols/L
18.0 mols x (98g/mol) = 1764 grams H2SO4/L solution.
The mass of 1L solution is
1.84 g/mL x 1000 mL = 1840 grams H2SO4 + H2O
%H2SO4 = (mass H2SO4/total mass)*100 = (1764/1840)*100 = ?
mols H2SO4 = 1764/98 = ?
mols H2O = (1840-1764)/18 = ?
total mols = sum
XH2SO4 = mols H2SO4/total mols
XH2O = mols H2O/total mols.
molality = mols/kg solvent. Just substitute the proper numbers.
Answer this Question
#1 250mL x 3.0 M/18.0M = 41.67mL?
#2 molality of solution???
Mass % 1764/1840x100= 95.87%
Mol fraction...H2SO4. 18/22.2=.81
H2O 4.2/22.2=.18
#1 looks ok.
#2.
mass % looks ok but if your prof is picky about the number of significant figures I think you need to round that number.
mol fractions look ok except I would have rounded the H2O to 0.19 and not 0.18. In fact, you dropped a 2 from 4.22 and 4.22/22.2 = 0.19
I don't see that you have calculated molality.
That's mols/kg solvent.
You have 18.0 mols and kg H2O = 0.076; therefore, m = 18.0/0.076 = ?
To make a 3.0 M solution of H2SO4 with a volume of 250.0 mL, you need to follow these steps:
Step 1: Determine the amount of H2SO4 needed
The equation for molality is molarity (M) = moles (mol) / kilograms (kg). In this case, you want to find the moles of H2SO4 needed. Use the equation:
moles of H2SO4 = molarity (M) x volume (L)
Convert the 250.0 mL volume to liters by dividing by 1000:
volume (L) = 250.0 mL / 1000 = 0.250 L
Now, substitute the values into the equation:
moles of H2SO4 = 3.0 M x 0.250 L = 0.750 mol
So, you need 0.750 moles of H2SO4 for the solution.
Step 2: Calculate the mass of H2SO4 needed
To calculate the mass of H2SO4 needed, you can use its molar mass. The molar mass of H2SO4 is:
2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol
Now, use the equation:
mass of H2SO4 = moles of H2SO4 x molar mass of H2SO4
Substitute the values:
mass of H2SO4 = 0.750 mol x 98.09 g/mol = 73.57 g
So, you need 73.57 grams of H2SO4 for the solution.
Step 3: Calculate the molality of the solution
The molality (m) is defined as the moles of solute divided by the mass of the solvent in kilograms. In this case, the moles of solute is 0.750 mol (calculated earlier). The mass of the solvent (water) can be calculated using its density.
mass of solution = volume of solution x density of solution
Substitute the values:
mass of solution = 250.0 mL x 1.84 g/mL = 460 g
Now, calculate the mass of the solvent (water):
mass of water = mass of solution - mass of H2SO4 = 460 g - 73.57 g = 386.43 g
Convert the mass of water to kilograms:
mass of water = 386.43 g / 1000 = 0.386 kg
Finally, substitute the values into the molality equation:
molality = moles of solute / mass of solvent (kg)
molality = 0.750 mol / 0.386 kg = 1.943 mol/kg
So, the molality of the solution is approximately 1.943 mol/kg.
To calculate the mass percent of H2SO4 in the solution, you can use the equation:
mass percent = (mass of H2SO4 / mass of solution) x 100
Substitute the values:
mass percent = (73.57 g / 460 g) x 100 = 15.99%
So, the mass percent of H2SO4 in the solution is approximately 15.99%.
The mole fraction (X) of H2SO4 in the solution can be calculated using the equation:
X = moles of H2SO4 / total moles
For the given 0.750 moles of H2SO4, the total moles will be the sum of moles of H2SO4 and moles of water. The moles of water can be calculated using its molar mass:
molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
mass of water = mass of solution - mass of H2SO4 = 460 g - 73.57 g = 386.43 g
moles of water = mass of water / molar mass of H2O = 386.43 g / 18.02 g/mol = 21.44 mol
Now, calculate the total moles:
total moles = moles of H2SO4 + moles of water = 0.750 mol + 21.44 mol = 22.19 mol
Finally, substitute the values into the mole fraction equation:
X = 0.750 mol / 22.19 mol = 0.0338
So, the mole fraction of H2SO4 in the solution is approximately 0.0338, and the mole fraction of water is approximately (1 - 0.0338) = 0.9662.