Question

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

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

Step-by-step solution

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.

Determine the volume of the solution

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

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.