Any help is appreciated

1.63×10¹⁰ L.

Explanation:

We'll begin by calculating the number of mole in 125 g of silver (ii) oxide, AgO. This can be obtained as follow:

Mass of AgO = 125 g

Molar mass of AgO = 108 + 16 = 124 g/mol.

Mole of AgO =?

Mole = mass /Molar mass

Mole of AgO = 125 /124

Mole of AgO = 1.01 moles.

Next, we shall convert 6.19×10¯⁵ μmol/L to mol/L. This can be obtained as follow:

1 μmol/L = 1×10¯⁶ mol/L

Therefore,

6.19×10¯⁵ μmol/L = 6.19×10¯⁵ × 1×10¯⁶ = 6.19×10¯¹¹ mol/L

Finally, we shall determine the volume as follow:

Molarity = 6.19×10¯¹¹ mol/L

Mole of AgO = 1.01 moles

Volume =?

Molarity = mole /Volume

6.19×10¯¹¹ = 1.01/Volume

Cross multiply

6.19×10¯¹¹ × volume = 1.01

Divide both side by 6.19×10¯¹¹

Volume = 1.01/6.19×10¯¹¹

Volume = 1.63×10¹⁰ L

Therefore, the volume of the solution is 1.63×10¹⁰ L