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Sodium acetate NaC2H3O2, has a formula weight of 82.034 grams g/mol. What is the molarity of a solution prepared by dissolving 47.6 g of sodium acetate in enough water to prepare 173.6mL of the solution?

Respuesta :

Solute: NaC2H3O2

The molar mass = 82.034 g/mol.

To calculate molarity we use the next formula:

M = number of moles of solute/volume of solution (L)

M = mass of solute/The molar mass x Volume (L)

Volume = 173.6 mL = 0.1736 L (1 L = 1000 mL)

M = 47.6 g/(82.034 g/mol x 0.1736 L) = 3.34 mol/L

Answer: M = 3.34 mol/L