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

(c) \(1.5 \times 10^{23}\)

Step-by-step solution

Understanding the Given Information

The boy drinks a 500 mL solution that has a glucose concentration of 9%. This means 9% of the solution's weight is glucose. We need to calculate the number of glucose molecules consumed.

Calculating the Mass of Glucose

To find the mass of glucose, we calculate 9% of the solution's volume. Since the solution's density is assumed to be similar to water (1 g/mL), the mass of the solution is also 500 g. Thus, the mass of glucose is \(\frac{9}{100} \times 500 = 45\) grams.

Calculating Moles of Glucose

The number of moles of glucose is the mass divided by the molar mass of glucose (180 g/mol). This gives us \(\frac{45}{180} = 0.25\) moles.

Converting Moles to Number of Molecules

To convert moles to molecules, we use Avogadro's number, \(6.022 \times 10^{23}\). Thus, the number of glucose molecules is \(0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}\) molecules.

Choosing the Closest Option

Comparing to the given options, \(1.5055 \times 10^{23}\) molecules is closest to choice (c), \(1.5 \times 10^{23}\).

Key concepts

Glucose Solution Concentration

When dealing with glucose solution concentration, it's all about understanding how much glucose is present in a given volume of solution. In this exercise, we work with a 9% glucose solution. This percentage tells us the proportion of glucose in the solution. In other words, 9% of the total mass of the solution is glucose. To make sense of this:

• If you have 100 grams of solution, 9 grams of it is glucose.
• In our case, with 500 mL of solution, assuming the density is like water and therefore 500 grams, 9% of the total weight is glucose, which calculates to 45 grams of glucose.

Understanding glucose concentration is crucial in fields such as chemistry and biology because it helps determine the exact amount of a chemical substance in a mixture. This knowledge is key when preparing solutions in laboratories and for conducting experiments accurately.
Determining concentrations helps in understanding how solutions will react chemically, and the efficiency of biological processes in applications like medical treatments and nutrition.

Avogadro's Number

Avogadro's number, which is approximately \(6.022 \times 10^{23}\), plays a pivotal role in chemistry as it helps convert between the macroscopic scale that we can observe and the microscopic scale of individual atoms and molecules. Named after the scientist Amedeo Avogadro, this constant represents the number of particles—whether atoms, molecules, or ions—in one mole of a substance.

When calculating the number of molecules in this glucose solution, knowing Avogadro's number allows us to bridge the gap between moles (a countable amount of substance) and the actual number of individual glucose molecules present.

• In our example, once we determined we had 0.25 moles of glucose, multiplying by Avogadro's number gave us the actual number of molecules: \(0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23}\).

Avogadro's number is foundational in understanding and balancing chemical reactions, calculating concentrations, and exploring the properties of gases, providing a common scale to work with in chemistry.

Molar Mass Calculations

The concept of molar mass connects the mass of a substance to the number of particles it contains, measured in grams per mole (g/mol). In this exercise, we focused on glucose, a common sugar with a molar mass of 180 g/mol. Molar mass is essential in converting between the mass of a substance and its amount in moles, facilitating quantitative calculations in chemistry.

• For any chemical substance: the molar mass is calculated by summing the masses of individual atoms indicated by the molecular formula.
• In glucose (\(C_6H_{12}O_6\)), adding six carbons, twelve hydrogens, and six oxygens gives us a molar mass of 180 g/mol.

Understanding molar mass is crucial when you need to convert a measured mass to the number of moles, allowing you to calculate the number of particles, as seen when calculating the number of glucose molecules. This concept is employed in stoichiometry, which involves the calculations of reactants and products in chemical reactions, ensuring precise predictions in laboratory and industrial processes.