How many liters of water are required to dissolve 1.00 g of silver chromate, given its solubility is 3.65×10^(-10) g/L?

• A. 1.00 × 10^10 L
• B. 2.74 × 10^9 L
• C. 5.00 × 10^9 L
• D. 3.65 × 10^8 L

Answer: (B)

Verified step by step guidance

1. Understand the problem: We need to find out how many liters of water are required to dissolve 1.00 g of silver chromate, given its solubility is 3.65×10^(-10) g/L.
2. Identify the relationship between solubility and the amount of solute: Solubility is defined as the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature. Here, it is given as 3.65×10^(-10) g/L.
3. Set up the equation: Use the formula for solubility, which is Solubility (g/L) = Mass of solute (g) / Volume of solvent (L). Rearrange this to find the volume of solvent: Volume of solvent (L) = Mass of solute (g) / Solubility (g/L).
4. Substitute the given values into the equation: Mass of solute = 1.00 g and Solubility = 3.65×10^(-10) g/L. So, Volume of solvent (L) = 1.00 g / 3.65×10^(-10) g/L.
5. Calculate the volume: Perform the division to find the volume of water required to dissolve 1.00 g of silver chromate. This will give you the answer in liters.