Question

Calculate the molarity of a solution that contains 85.0 g of Zn(C2H3O2)2 in 250. mL of solution (don't forget to convert mL to L first). Round to the nearest hundredth.

Answer 1

Answer:

[Zn(C₂H₃O₂)₂] = 1.85M (3 sig-figs accurate to 0.01 Molar)

Explanation:

Concentration is defined as amount of solute in a specified volume of solution. That is, Concentration = mass of solute/volume of solution (solution = solute + solvent). When concentration is in terms of Molar values the essential relationship is Molarity(M) = moles solute / volume of solution in liters.

For this problem => first, convert mass of Zn(C₂H₃O₂)₂ into moles and then the volume of solution into liters. Take ratio of moles/volume of solution (L).

=> Molar Concentration = moles of Zn(C₂H₃O₂)₂ / Liters of solution

moles of Zn(C₂H₃O₂)₂ = 85.0 grams Zn(C₂H₃O₂)₂ / formula weight of Zn(C₂H₃O₂)₂  =  85.0g/183.5g·mole⁻¹ = 0.463 mole Zn(C₂H₃O₂)₂.

Volume of solution in Liters = 250ml / 1000ml/L = 0.250 liters

Molarity of  Zn(C₂H₃O₂)₂ solution = 0.463 mole Zn(C₂H₃O₂)₂  /0.250 liters of solution = 1.85 Molar in Zn(C₂H₃O₂)₂

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Note: The symbiology convention for 'molar solution concentration' is to place brackets around the molecular formula. That is, [Zn(C₂H₃O₂)₂] = 1.85M