Problem 16 - Chemistry Matter and Change

Question

What is the molarity of an aqueous solution containing 40.0 \(\mathrm{g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in 1.5 \(\mathrm{L}\) of solution?

Step-by-Step Solution

Answer

The molarity of the aqueous solution containing 40.0 g of glucose in 1.5 L of solution is 0.148 M.

Step 1: Calculate the molar mass of glucose (C6H12O6)

First, let's calculate the molar mass of glucose, using the atomic masses of carbon (C), hydrogen (H), and oxygen (O) found on a periodic table. Glucose has 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Molar mass of glucose = (6 * molar mass of C) + (12 * molar mass of H) + (6 * molar mass of O) Molar mass of glucose = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) Molar mass of glucose = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol Molar_mass of glucose = 180.18 g/mol

Step 2: Convert the mass of glucose to moles

Now we can convert the 40.0 g of glucose into moles, using the molar mass we calculated above: moles of glucose = (mass of glucose) / (molar mass of glucose) moles of glucose = (40.0 g) / (180.18 g/mol) moles of glucose = 0.222 mol

Step 3: Calculate the molarity of the solution

Now that we know the moles of glucose, we can calculate the molarity of the solution using the formula: Molarity = (moles of solute) / (volume of solution in liters) Molarity = (0.222 mol) / (1.5 L) Molarity = 0.148 M So, the molarity of the aqueous solution containing 40.0 g of glucose in 1.5 L of solution is 0.148 M.

Key Concepts

Molar Mass

Understanding molar mass is fundamental when dealing with chemical calculations. In simple terms, molar mass is the mass of one mole of a substance. A mole is a handy unit used in chemistry to express amounts of a chemical substance. To calculate the molar mass of a compound like glucose, we sum the atomic masses of all the atoms present in the molecule.

For glucose \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \), this involves adding the mass of:

• 6 carbon atoms
• 12 hydrogen atoms
• 6 oxygen atoms

Consult a periodic table to determine the atomic masses:

• \( \mathrm{Carbon} = 12.01 \, \mathrm{g/mol} \)
• \( \mathrm{Hydrogen} = 1.01 \, \mathrm{g/mol} \)
• \( \mathrm{Oxygen} = 16.00 \, \mathrm{g/mol} \)

By multiplying the number of each type of atom with its atomic mass and then adding them together, we find that the molar mass of glucose is:\( 72.06 \, \mathrm{g/mol} + 12.12 \, \mathrm{g/mol} + 96.00 \, \mathrm{g/mol} = 180.18 \, \mathrm{g/mol} \).By knowing these calculations, you can understand how much one mole of glucose weighs, which is vital for converting between grams and moles in calculations.

Glucose Calculation

Calculating the number of moles from a given mass is a common task in chemistry. In this example, you have 40.0 grams of glucose. From the previous section, we know the molar mass of glucose is \( 180.18 \ \, \mathrm{g/mol} \).

To find the moles of glucose, use the formula:\[\text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}.\]Plugging in your values gives:\[\text{Moles of glucose} = \frac{40.0 \, \mathrm{g}}{180.18 \, \mathrm{g/mol}} \approx 0.222 \, \text{mol}.\]This tells you that there are approximately 0.222 moles of glucose in 40.0 grams. Knowing how to do this conversion is key to determining the amount of substance you're dealing with in moles.

Chemical Solution Concentration

Molarity is an important concept when discussing the concentration of a solution. It tells us how much solute (the substance being dissolved) is present in a given volume of solution. This is calculated in moles per liter, giving a unit of molarity (symbol: \( M \)).

For the glucose solution, you have calculated that there are 0.222 moles of glucose in a 1.5-liter solution. To determine the molarity, use this simple formula:\[\text{Molarity} \, (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}.\]Substituting the known values provides:\[M = \frac{0.222 \, \text{mol}}{1.5 \, \text{L}} \approx 0.148 \, M.\]This result means there are 0.148 moles of glucose in every liter of solution. Understanding molarity helps in comprehending how concentrated a solution is, which is crucial in various scientific and practical applications.