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Chapter 11: Problem 11

Rubbing alcohol contains 585 g isopropanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) per liter (aqueous solution). Calculate the molarity.

Short Answer

Expert verified
The molarity of the isopropanol in the rubbing alcohol solution is 9.74 M.

Step by step solution

01

Convert mass to moles

First, we need to find the molar mass of isopropanol (C₃H₇OH). The molar mass of C=12.01 g/mol, H=1.008 g/mol, and O=16.00 g/mol. The molecular formula of isopropanol is C₃H₈O, so its molar mass is: (3 * 12.01) + (8 * 1.008) + (1 * 16.00) = 36.03 + 8.064 + 16.00 ≈ 60.10 g/mol. Now, we'll convert the mass of isopropanol (585 g) into moles by using the molar mass: moles of isopropanol = mass / molar mass = 585 g / 60.10 g/mol ≈ 9.74 mol.
02

Determine the volume of the solution

We are given that the solution has a volume of 1 liter.
03

Calculate the molarity

Now that we have the moles of isopropanol and the volume of the solution, we can calculate the molarity: Molarity (M) = moles of solute / volume of solution in liters = 9.74 mol / 1 L = 9.74 M. So, the molarity of the isopropanol in the rubbing alcohol solution is 9.74 M.

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Most popular questions from this chapter

Explain the terms isotonic solution, crenation, and hemolysis.

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