A. A solution of rubbing alcohol is 77.7% (V/W) isopropanol in water. How many milliliters of isopropanol are in an 82.7 mL sample of the rubbing alcohol solution?
Express the answer to 3 significant figures.

B. How many liters of a 3.64 M K2SO4 solution are needed to provide 98.2 g of K2SO4 (molar mass 174.01 g/mol)?
Recall that M is equivalent to mol/L.
Express the answer to 3 significant figures.

A- The sample contains approximately 64.1 mL of isopropanol. B- Approximately 0.563 L of a 3.64 M K2SO4 solution is needed to provide 98.2 g of K2SO4.

Explanation:

Calculating the amount of isopropanol in the rubbing alcohol solution:

To calculate the amount of isopropanol in the given sample, we can use the formula:

Amount of isopropanol = (Percentage of isopropanol / 100) * Volume of the solution

Given that the percentage of isopropanol is 77.7% and the volume of the solution is 82.7 mL, we can substitute these values into the formula:

Amount of isopropanol = (77.7 / 100) * 82.7 mL

Calculating this expression gives us:

Amount of isopropanol = 0.777 * 82.7 mL = 64.1 mL

Calculating the volume of a 3.64 M K2SO4 solution:

To calculate the volume of the K2SO4 solution required, we can use the formula:

Volume of solution = Mass of solute / Molar mass * Molarity

Given that the mass of K2SO4 is 98.2 g, the molar mass of K2SO4 is 174.01 g/mol, and the molarity of the solution is 3.64 M, we can substitute these values into the formula:

Volume of solution = 98.2 g / 174.01 g/mol * 3.64 M

Calculating this expression gives us:

Volume of solution = 0.563 L

Learn more about calculating the amount of a substance in a solution here:

https://brainly.com/question/8087376

#SPJ14