Question

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?

Answer 1

Ce concentration = 0.0387 M

Step-by-step explanation:

The given compound is: (NH₄)₂Ce(NO₃)₆

Mass = 10.6192 g

Mol. wt = 548.23 g/mol

`Moles = (Mass)/(Mol.wt)=(10.6192g)/(548.23g/mol)=0.01937`

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

`Molarity = (Moles)/(Volume)=(0.01937moles)/(0.500L)=0.0387 M`