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Chapter 11: Problem 11

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

Short Answer

Expert verified
The molar mass of isopropanol is \(M_\text{C3H7OH} = 3(12.01) + 8(1.01) + 1(16.00) = 60.10 \text{ g/mol}\). Next, calculate the moles of isopropanol: \(\text{moles} = \frac{585 \text{ g}}{60.10 \text{ g/mol}} = 9.74 \text{ mol}\). Finally, calculate the molarity: \(\text{Molarity} = \frac{9.74 \text{ mol}}{1 \text{ L}} = 9.74 \text{ M}\).

Step by step solution

01

Determine the molar mass of isopropanol

The first step is to determine the molar mass of isopropanol, C3H7OH. The formula contains 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom. Using the periodic table, we can find the atomic masses of carbon (C), hydrogen (H), and oxygen (O) as follows: \[M_\text{C} = 12.01 \text{ g/mol}\] \[M_\text{H} = 1.01 \text{ g/mol}\] \[M_\text{O} = 16.00 \text{ g/mol}\] Now, we can calculate the molar mass of isopropanol: \[M_\text{C3H7OH} = 3M_\text{C} + 8M_\text{H} + 1M_\text{O}\]
02

Calculate the moles of isopropanol

Next, we need to calculate the number of moles of isopropanol in the solution. We are given that there are 585 g of isopropanol in 1 L of the solution. We can convert this mass into moles using the molar mass of isopropanol: \[\text{moles of isopropanol} = \frac{\text{mass of isopropanol}}{\text{molar mass of isopropanol}}\]
03

Calculate the molarity

Finally, we can calculate the molarity of the solution by dividing the number of moles of isopropanol by the volume of the solution in liters: \[\text{Molarity} = \frac{\text{moles of isopropanol}}{\text{volume in liters}}\] We already have the number of moles (from step 2) and the volume (1 L) of the solution. Simply substitute the values and calculate the molarity.

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