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What is the mass of zinc acetate (183.49 g/mol) that is dissolved in 400 ml of 0.500 M Zn(C₂H₃O₂)2 solution?

User Fazila
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1 Answer

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Final answer:

The mass of zinc acetate dissolved in 400 ml of 0.500 M Zn(C₂H₃O₂)₂ solution is determined to be 36.698 g by multiplying the number of moles (0.200 mol) by the molar mass (183.49 g/mol).

Step-by-step explanation:

To find the mass of zinc acetate dissolved in a 0.500 M solution, first, determine the number of moles contained in 400 ml of this solution:




Number of moles = Molarity × Volume (in liters)
Number of moles = 0.500 mol/L × 0.400 L
Number of moles = 0.200 mol



To calculate the mass of zinc acetate, use the formula:




Mass = Number of moles × Molar mass
Mass = 0.200 mol × 183.49 g/mol
Mass = 36.698 g



Therefore, the mass of zinc acetate dissolved in 400 ml of 0.500 M Zn(C₂H₃O₂)2 solution is 36.698 g.

User Dan Gravell
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