Calculate the number of moles of ag+ in 5.00 ml of 2.00×10−3 m agno3 and the number of moles of cro2−4 in 5.00 ml of 3.10×10−3 m k2cro4.

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

Abbreviated as mg-25.