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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)

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)

b. L ≈ 4.00 litres

c. L ≈ 3.4 litres

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

it would be a gas formation