Solution Found!
Gold occurs in seawater at an average concentration of 1.1
Chapter 14, Problem 85P(choose chapter or problem)
Problem 85P
Gold occurs in seawater at an average concentration of 1.1 × 10?2 ppb. How many liters of seawater must be processed to recover 1 troy ounce of gold, assuming 81.5% efficiency (d of seawater = 1.025 g/mL; 1 troy ounce = 31.1 g)?
Questions & Answers
QUESTION:
Problem 85P
Gold occurs in seawater at an average concentration of 1.1 × 10?2 ppb. How many liters of seawater must be processed to recover 1 troy ounce of gold, assuming 81.5% efficiency (d of seawater = 1.025 g/mL; 1 troy ounce = 31.1 g)?
ANSWER:
Solution 85P
Here, we are going to determine the volume of seawater.
Step 1:
Given that,
The concentration of gold = 1.1 × 10-2 ppb
Density of the sea water = 1.025 g/mL
We know that,
----(1)
Calculation of the mass of solvent:
= 1.025 g/mL ----(2)