Respuesta :
Molarity is defined as the number of moles of solute in 1 L of solution
the mass of sodium acetate added - 12.0 g
molar mass of sodium acetate - 82 g/mol
number of moles = mass present / molar mass
number of moles of sodium acetate added = 12.0 g / 82 g/mol = 0.146 mol
the number of moles in 250.0 mL - 0.146 mol
therefore number of moles in 1000 mL - 0.146 mol / 250.0 mL x 1000 mL/L = 0.584 mol
there are 0.584 mol in 1 L - 0.584 mol/L
the molarity of sodium acetate is 0.584 M
the mass of sodium acetate added - 12.0 g
molar mass of sodium acetate - 82 g/mol
number of moles = mass present / molar mass
number of moles of sodium acetate added = 12.0 g / 82 g/mol = 0.146 mol
the number of moles in 250.0 mL - 0.146 mol
therefore number of moles in 1000 mL - 0.146 mol / 250.0 mL x 1000 mL/L = 0.584 mol
there are 0.584 mol in 1 L - 0.584 mol/L
the molarity of sodium acetate is 0.584 M
The molarity of the solution prepared by dissolving 12.0 g of sodium acetate is 0.584M.
How we calculate molarity?
Molarity of any solution is define as the no. of moles of solute present in per liter of solution, and for this question moles of sodium acetate will be calculated as :
n = W/M, where
W = given mass = 12g
M = molar mass = 82 g/mol
Moles of sodium acetate = 12/82 = 0.146 mole
Given volume of water = 250mL = 0.250L
Molarity of solution = 0.146mol / 0.250L = 0.584M
Hence, 0.584M is the molarity of the solution.
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