We will follow the hint. First we will find the number of moles necessary to make 2.2 L of a 0.45 M solution of sodium bicarbonate. Then we will convert those moles to mass.
The molarity is defined like:
Molarity = number of moles of solute/(volume of solution in L)
We know the molarity and volume of solution in L.
Molarity = 0.45 M
Volume of solution in L = 2.2 L
The number of moles of sodium bicarbonate are:
Molarity = number of moles of solute/(volume of solution in L)
number of moles of solute = molarity * volume of solution in L
number of moles of solute = 0.45 M * 2.2 L
number of moles of solute = 0.99 moles
moles of NaHCO₃ = 0.99 moles
Now we can find the mass of sodium bicarbonate that we have in 0.99 moles of it. We will use the molar mass of sodium bicarbonate.
formula of sodium bicarbonate: NaHCO₃
atomic mass of Na = 22.99 amu
atomic mass of H = 1.01 amu
atomic mass of C = 12.01 amu
atomic mass of O = 16.00 amu
molar mass of NaHCO₃ = 22.99 + 1.01 + 12.01 + 3 * 16.00
molar mass of NaHCO₃ = 84.01 g/mol
mass of NaHCO₃ = moles of NaHCO₃ * molar mass of NaHCO₃
mass of NaHCO₃ = 0.99 moles * 84.01 g/mol
mass of NaHCO₃ = 83.2 g
Answer: 83.2 g of sodium bicarbonate is need to make the solution.