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

= \( 36.03 + 7.07 + 16.00 = 59.10 \: \mathrm{g/mol} \) #tag_title# Step 2: Determine the number of moles of isopropanol #tag_content# We know that there are 585 g of isopropanol in 1 liter of solution. To find the number of moles, we will use the formula: Moles = \( \frac{mass}{molar \: mass} \) Moles = \( \frac{585 \: \mathrm g}{59.10 \: \mathrm{g/mol}} = 9.90 \: \mathrm{moles} \) #tag_title# Step 3: Calculate the molarity #tag_content# Molarity is the number of moles of solute per liter of solution. Since we have 9.90 moles of isopropanol in 1 liter of solution, the molarity of this solution is: Molarity = \( \frac{9.90 \: \mathrm{moles}}{1 \: \mathrm L} = 9.90 \: \mathrm{M} \) The molarity of the rubbing alcohol solution is 9.90 M.