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A standard solution is prepared by dissolving 10.6192 g of (NH4)2Ce(NO3)6 (548.23 g•mol-1, 98.75% purity) in dilute sulfuric acid. The resulting solution is quantitatively transferred to a 500.0-mL volumetric flask and diluted to the mark. What is the Ce concentration in the final solution?

Respuesta :

Answer:

Ce concentration = 0.0387 M

Explanation:

The given compound is: (NH₄)₂Ce(NO₃)₆

Mass = 10.6192 g

Mol. wt = 548.23 g/mol

[tex]Moles = \frac{Mass}{Mol.wt}=\frac{10.6192g}{548.23g/mol}=0.01937[/tex]

Based on the formula stoichiometry:

1 mole of (NH₄)₂Ce(NO₃)₆ contains 1 mole of Ce

Therefore, there are 0.01937 moles of Ce in the given compound

Volume of the solution = 500.0 ml = 0.500 L

[tex]Molarity = \frac{Moles}{Volume}=\frac{0.01937moles}{0.500L}=0.0387 M[/tex]

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