I'm going crazy please help me out
part 1 of 2 Determine the mass of anhydrous copper(II) sulfate that must be used to prepare 250 mL of 2.01 M CuSO4(aq). Which i found to be 80.204g
part 2 of 2
Determine the mass of CuSO4.5H2O that must be used to prepare 250 mL of 2.01 M CuSO4(aq).
another question
Na3AsO4 is a salt of a weak base that can accept more than one proton. If 30.2 g of Na3AsO4 is dissolved in water to make 250 mL of solution, how many moles of sodium cations are in the solution?
Answer in moles
1. 80.204 is ok except you are not allowed that many significant figures.
2. This one is done just like #1 but instead of molar mass being 159.61 it is 159.61+(5*80)= ?? or about that. YOu need to check that out.
3. moles Na3AsO4 = grams/molar mass.
moles Na^+ = 3x that. The amount of water in which it is places makes no difference.
of 1.79 M CuSO4(aq).
To determine the mass of CuSO4.5H2O that must be used to prepare 250 mL of 2.01 M CuSO4(aq), you need to consider the molar mass of CuSO4.5H2O and the molarity of the solution.
1. Determine the molar mass of CuSO4.5H2O:
Cu: 1 atom x atomic mass of Cu
S: 1 atom x atomic mass of S
O: 4 atoms x atomic mass of O
H2O: 5 molecules x (2 atoms x atomic mass of H + 1 atom x atomic mass of O)
The atomic masses are:
Cu: 63.55 g/mol
S: 32.06 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Calculate the molar mass:
Molar mass of CuSO4.5H2O = (1 x 63.55 g/mol) + (1 x 32.06 g/mol) + (4 x 16.00 g/mol) + (5 x [(2 x 1.01 g/mol) + 16.00 g/mol])
2. Use the formula for molarity (M) to calculate moles of CuSO4 in the solution:
M = moles of solute / volume of solution (in liters)
Rearrange the formula to solve for moles of solute:
moles of solute = M x volume of solution (in liters)
In this case, the molarity (M) is 2.01 M, and the volume of solution is 250 mL (0.250 L).
moles of solute = 2.01 M x 0.250 L
3. Finally, calculate the mass of CuSO4.5H2O using the moles of solute and the molar mass of CuSO4.5H2O:
mass = moles of solute x molar mass of CuSO4.5H2O
Substitute the values:
mass = (2.01 M x 0.250 L) x molar mass of CuSO4.5H2O
By solving this equation using the molar mass of CuSO4.5H2O, you will find the mass required to prepare 250 mL of the 2.01 M CuSO4(aq) solution.
For the second question, the formula Na3AsO4 indicates that there are three sodium ions (Na+) per formula unit. Therefore, if you have 30.2 g of Na3AsO4 dissolved in 250 mL of solution, you can calculate the molarity of sodium cations using the molar mass of Na3AsO4 and the volume of the solution.
1. Determine the molar mass of Na3AsO4:
Na: 3 atoms x atomic mass of Na
As: 1 atom x atomic mass of As
O: 4 atoms x atomic mass of O
The atomic masses are:
Na: 22.99 g/mol
As: 74.92 g/mol
O: 16.00 g/mol
Calculate the molar mass:
Molar mass of Na3AsO4 = (3 x 22.99 g/mol) + (1 x 74.92 g/mol) + (4 x 16.00 g/mol)
2. Calculate the moles of Na3AsO4 in the solution:
moles = mass / molar mass
Substitute the given values:
moles = 30.2 g / molar mass of Na3AsO4
3. Determine the moles of sodium cations (Na+):
Since there are three sodium ions per formula unit of Na3AsO4, you can multiply the moles of Na3AsO4 by three to obtain the moles of sodium cations.
moles of sodium cations = 3 x moles of Na3AsO4
By substituting the values, you will find the moles of sodium cations in the solution. Note that the final answer should be in moles.
To determine the mass of CuSO4·5H2O needed to prepare 250 mL of a 2.01 M CuSO4(aq) solution, we need to consider the molar mass and stoichiometry.
Step 1: Calculate the number of moles of CuSO4 needed.
We know that the molarity (M) is given as 2.01 moles/L. Since we have 250 mL of solution (0.250 L), we can calculate the number of moles required using the formula:
moles = Molarity x Volume
moles = 2.01 mol/L x 0.250 L
moles = 0.5025 mol
Step 2: Determine the molar mass of CuSO4·5H2O.
CuSO4·5H2O is a hydrated compound, which means it contains water molecules in addition to the anhydrous salt CuSO4. The molar mass of anhydrous CuSO4 is approximately 159.61 g/mol, and the molar mass of water (H2O) is about 18.015 g/mol.
To calculate the molar mass of CuSO4·5H2O, we sum the molar masses of the individual elements:
Molar mass = (mass of Cu) + (mass of S) + (4 x mass of O) + (5 x mass of water)
Molar mass = (63.546 g/mol) + (32.06 g/mol) + (4 x 16.00 g/mol) + (5 x 18.015 g/mol)
Molar mass = 249.69 g/mol
Step 3: Calculate the mass of CuSO4·5H2O needed.
To find the mass of CuSO4·5H2O, we use the following equation:
mass = moles x molar mass
mass = 0.5025 mol x 249.69 g/mol
mass ≈ 125.50 g
Therefore, the mass of CuSO4·5H2O needed to prepare 250 mL of 2.01 M CuSO4(aq) solution is approximately 125.50 g.
For the second question about Na3AsO4, we need to determine the number of moles of sodium cations (Na+) in the solution.
Step 1: Calculate the molar mass of Na3AsO4.
The molar mass of Na3AsO4 can be determined by summing the molar masses of sodium (Na), arsenic (As), and oxygen (O). The atomic masses of these elements are approximately:
Na: 22.99 g/mol
As: 74.92160 g/mol
O: 16.00 g/mol
Molar mass = (3 x mass of Na) + mass of As + (4 x mass of O)
Molar mass = (3 x 22.99 g/mol) + 74.92160 g/mol + (4 x 16.00 g/mol)
Molar mass ≈ 201.93 g/mol
Step 2: Calculate the number of moles of Na3AsO4 dissolved.
To calculate the number of moles of Na3AsO4, we use the formula:
moles = mass / molar mass
moles = 30.2 g / 201.93 g/mol
moles ≈ 0.1498 mol
Step 3: Determine the number of moles of sodium cations.
Since Na3AsO4 has three sodium ions (Na+) per formula unit, the number of moles of sodium cations is three times the number of moles of Na3AsO4:
moles of Na+ = 3 x moles of Na3AsO4
moles of Na+ ≈ 3 x 0.1498 mol
moles of Na+ ≈ 0.4494 mol
Therefore, there are approximately 0.4494 moles of sodium cations in the solution.