Answer :
The mass of NaNO₃ required to prepare 305 ml of a 0.577 M solution of NaNO₃ is 14.957 g.
Concentration, or the amount of solute in a solution, can either be expressed in molarity, M, moles of solute per liter of solution, or molality, m, moles of solute per kilogram of solution.
Molarity = moles of solute / liter of solution
Given the molar mass of NaNO₃, the number of moles of NaNO₃ can be calculated by dividing the mass of NaNO₃ by its molar mass.
moles of NaNO₃ = mass of NaNO₃ / molar mass of NaNO₃
moles of NaNO₃ = mass of NaNO₃ / 84.99 g/mol
Plug in the values to the equation for the molarity and solve for the required mass of NaNO₃.
Molarity = moles of solute / liter of solution
0.577 M = (mass of NaNO₃ / 84.99 g/mol) / 0.305 L
0.577 M (0.305 L) = (mass of NaNO₃ / 84.99 g/mol)
mass of NaNO₃ = 0.577 M (0.305 L) (84.99 g/mol)
mass of NaNO₃ = 14.957 g
Learn more about molarity here: brainly.com/question/23243759
#SPJ4
