complete question:
Calculate the volume of each solution, in liters.
b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl₂)
c. a 0.4 M solution containing 290 g of aluminum nitrate (Al(NO₃)₃)
Answer:
b. L ≈ 4.00 litres
c. L ≈ 3.4 litres
Step-by-step explanation:
Molarity is the number of moles of solute per litre of solution. Together a solute and a solvent makes a solution. The formula for molarity can be represented below
M = number of moles of the solute(mol)/L
where
M = molarity
mol = number of moles of the solute
L = litre of the solution.
Therefore ,
The volume of each solution can be computed below
b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl₂)
M = number of moles of the solute(mol)/L
number of moles of MnCl₂ = mass/molar mass
molar mass of MnCl₂ = 55 + 71 = 126 g/mol
number of moles of MnCl₂ = 250/126
number of moles of MnCl₂ = 1.9841 moles
0.5 = 1.9841/L
cross multiply
0.5L = 1.9841
L = 1.9841/0.5
L = 3.97
L ≈ 4.00 litres
c. a 0.4 M solution containing 290 g of aluminum nitrate Al(NO₃)₃
M = number of moles of the solute(mol)/L
number of moles = mass/molar mass
molar mass of Al(NO₃)₃ = 27 + 14 × 3 + 48 × 3 = 27 + 42 + 144 = 213
number of moles = 290/213 = 1.3615 moles
M = number of moles of the solute(mol)/L
0.4 = 1.3615/L
cross multiply
0.4L = 1.3615
divide both sides by 0.4
L = 1.3615/0.4
L = 3.4 litres