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------------------------------------------------ Calculate the number of moles of [tex]\text{Ag}^+[/tex] in 5.00 mL of [tex]2.00 \times 10^{-3} \, M[/tex] [tex]\text{AgNO}_3[/tex] and the number of moles of [tex]\text{CrO}_4^{2-}[/tex] in 5.00 mL of [tex]3.10 \times 10^{-3} \, M[/tex] [tex]\text{K}_2\text{CrO}_4[/tex].

Answer :

Molarity = number of moles of solute / Volume of solution

the unit : mol / L

For AgNO₃ solution,
AgNO₃(aq) →Ag⁺(aq) + NO₃⁻(aq)
since stoichiometric ratio between AgNO₃(aq) and Ag⁺(aq) is 1 : 1,
the molarity of Ag⁺(aq) is 2.00×10⁻³ M
Hence moles of Ag⁺(aq) in 5.00 mL = (2.00×10⁻³ mol / L ) * 5.00 x 10⁻³ L
= 1.00 x 10⁻⁵ mol
For K₂CrO₄ solution,
K₂CrO₄(aq) →2K⁺(aq) + CrO₄²⁻(aq)
since stoichiometric ratio between K₂CrO₄(aq) and CrO₄²⁻(aq) is 1 : 1,
the molarity of CrO₄²⁻(aq) is 3.10×10⁻³ M
Hence moles of CrO₄²⁻(aq) in 5.00 mL = (3.10×10⁻³ mol / L ) * 5.00 x 10⁻³ L
= 1.55 x 10⁻⁵ mol

The number of moles of Ag⁺ is 1.00 x 10⁻⁵ and for CrO₄⁻² is 1.55 x 10⁻⁵.

Step-by-Step Calculation:

1. For Ag⁺ ions from AgNO₃:

  • Given: Volume = 5.00 mL (or 0.005 L), Molarity = 2.00 × 10⁻³ M

Using the formula: moles = molarity × volume

  • Moles of Ag⁺ = (2.00 × 10⁻³ M) × (0.005 L) = 1.00 × 10⁻⁵ moles

2. For CrO₄²⁻ ions from K₂CrO₄:

  • Given: Volume = 5.00 mL (or 0.005 L), Molarity = 3.10 × 10⁻³ M

Using the formula: moles = molarity × volume

  • Moles of CrO₄²⁻ = (3.10 × 10⁻³ M) × (0.005 L) = 1.55 × 10⁻⁵ moles

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