Question

A boy drinks \(500 \mathrm{~mL}\) of \(9 \%\) glucose solution. The number of glucose molecules he has consumed are [mol. wt. of glucose \(=180]\). (a) \(0.5 \times 10^{23}\) (b) \(1.0 \times 10^{23}\) (c) \(1.5 \times 10^{23}\) (d) \(2.0 \times 10^{23}\)

Short answer

The boy consumed approximately \(1.5 \times 10^{23}\) glucose molecules, so the correct answer is \((c) 1.5 \times 10^{23}\).

Step-by-step solution

Determine the mass of glucose in the solution

Since the solution is a \(9\%\) glucose solution, this means 9 grams of glucose are present in every 100 mL of solution. Therefore, in 500 mL of solution, the mass of glucose is calculated as follows: \( \frac{9 ext{ g}}{100 ext{ mL}} \times 500 ext{ mL} = 45 ext{ g}\).

Calculate the number of moles of glucose

To find the number of moles of glucose, use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of glucose is given as 180 g/mol. Therefore, the number of moles of glucose is \( \frac{45 ext{ g}}{180 ext{ g/mol}} = 0.25 ext{ moles} \).

Calculate the number of glucose molecules

To find the number of molecules, use Avogadro's number \(6.022 \times 10^{23}\) molecules/mol. Multiply the number of moles by Avogadro's number: \(0.25 ext{ moles} \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}\) molecules.

Choose the correct option

The calculated number of glucose molecules \(1.5055 \times 10^{23}\) is approximately \(1.5 \times 10^{23}\). Therefore, the correct answer is \((c) 1.5 \times 10^{23}\).

Key concepts

Molar Mass Calculation

Understanding molar mass is essential in chemistry because it helps us find out how much of a substance we have. Molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). For glucose, the molar mass is 180 g/mol. This means if you have one mole of glucose, it weighs 180 grams.
Calculating molar mass involves adding up the atomic masses of all atoms in a molecule. For glucose, the chemical formula is \( \text{C}_6\text{H}_{12}\text{O}_6 \). To calculate molar mass:

• Carbon (C) has an atomic mass of about 12; for 6 carbon atoms, that's \( 6 \times 12 = 72 \).
• Hydrogen (H) has an atomic mass of about 1; for 12 hydrogen atoms, that's \( 12 \times 1 = 12 \).
• Oxygen (O) has an atomic mass of about 16; for 6 oxygen atoms, that's \( 6 \times 16 = 96 \).

Adding these up gives \( 72 + 12 + 96 = 180 \) g/mol. Once you know the molar mass, you can find the number of moles of a substance by dividing its mass by its molar mass. This step is crucial in many chemical calculations.

Avogadro's Number

Avogadro's number connects the macroscopic world we can see with the microscopic world of atoms and molecules. It tells us how many atoms or molecules are in one mole of a substance, a huge number: \( 6.022 \times 10^{23} \).
This number might seem abstract, but it's essential for converting between moles and number of molecules. For example, if you have 0.25 moles of glucose, you can calculate the number of glucose molecules by multiplying the moles by Avogadro's number:

• Number of molecules = \( 0.25 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \)
• This equals \( 1.5055 \times 10^{23} \text{ molecules} \).

Understanding Avogadro's number is key to the mole concept, as it provides a link between the mass of a substance and how many individual molecules or atoms it contains.

Concentration Percentage

Concentration percentage is a way of expressing how much solute is present in a solution. In this case, it's expressed as a percentage of the total solution. For a \( 9\% \) glucose solution, this means that there are 9 grams of glucose for every 100 milliliters of the solution.
When you have a different volume of the solution, you can use this percentage to figure out how much solute is there. For example, in 500 mL of a \( 9\% \) glucose solution:

• First, use the concentration to determine the mass: \( \frac{9 \text{ g}}{100 \text{ mL}} \times 500 \text{ mL} = 45 \text{ g}\).
• This calculation tells you that there are 45 grams of glucose in 500 mL of the solution.

Knowing concentration percentages helps in calculating how much solute you've got, which is important for subsequent calculations involving the moles of solute and the number of molecules or atoms present.