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Ch.13 - Properties of Solutions
All textbooksBrown 14th EditionCh.13 - Properties of SolutionsProblem 96
Chapter 13, Problem 96

The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in 1.0 * 10^3 gal of seawater.

Verified step by step guidance
1
Convert the volume of seawater from gallons to liters using the conversion factor: 1 gallon = 3.78541 liters.
Calculate the mass of seawater in kilograms, assuming the density of seawater is approximately 1.025 kg/L.
Determine the mass of gold in the seawater using the concentration of 13 ppt (parts per trillion), which means 13 grams of gold per 10^12 grams of seawater.
Use the mass of seawater calculated in step 2 to find the mass of gold by setting up a proportion based on the ppt concentration.
Convert the mass of gold from grams to the desired unit if necessary, ensuring the final answer is in grams.
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Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.826 g/cm3. Calculate the concentration of the solution in (a) molality,

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Textbook Question

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is 0.826 g/cm3. Calculate the concentration of the solution in (b) mole fraction of LiBr,

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