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 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
