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

= \( 36.03 + 7.07 + 16.00 = 59.10 \: \mathrm{g/mol} \) #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.

Step-by-step solution

Find the molar mass of isopropanol (C3H7OH)

First, we have to find the molar mass of isopropanol by adding the molar mass of its individual atoms (3 carbon atoms, 7 hydrogen atoms, and 1 oxygen atoms). The molar masses are as follows: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Oxygen (O): 16.00 g/mol Now, let's calculate the molar mass of isopropanol: Molar mass = (3 * 12.01) + (7 * 1.01) + (1 * 16.00)