What is the molarity of an aqueous solution that is \(6.75 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass? (Assume a density of \(1.03 \mathrm{~g} / \mathrm{mL}\) for the solution.) (Hint: \(6.75 \% \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) by mass means \(6.75 \mathrm{~g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} / 100.0 \mathrm{~g}\) solution.)

Understanding the molar mass of glucose is essential when performing calculations involving chemical solutions. The molar mass is the weight of one mole of a substance, and it is typically expressed in grams per mole (g/mol).

For glucose, with the chemical formula \( C_6H_{12}O_6 \), the molar mass is calculated by adding up the atomic masses of all the atoms in a molecule of glucose. This involves six carbon (C) atoms, twelve hydrogen (H) atoms, and six oxygen (O) atoms. With the atomic weights given by the periodic table (\(12.01 \text{g/mol}\) for C, \(1.01 \text{g/mol}\) for H, and \(16.00 \text{g/mol}\) for O), the calculation becomes a straightforward summation.
\[ \text{Molar mass of glucose} = (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) \text{ g/mol} \]
Reaching a total of \(180.18 \text{g/mol}\) for the molar mass of glucose.

When dealing with chemical solutions, it's crucial to be able to convert between grams and moles, a process we call mole conversion. It enables us to understand how many molecules of a substance we have in a given mass.

This conversion uses the molar mass as a conversion factor. To convert the mass of glucose to moles, you divide the mass by the molar mass: \
\[ \text{moles glucose} = \frac{\text{mass (in grams)}}{\text{molar mass (in g/mol)}} \]
By applying this formula to our problem, \(6.75 \text{g}\) of glucose is equivalent to \(0.03746 \text{mol}\) when divided by the molar mass of \(180.18 \text{g/mol}\). Mole conversion allows us to work with the quantities involved in reactions or solutions at the molecular level, which is fundamental in chemistry.

Density is a property that compares the mass of a substance to its volume, usually expressed as grams per milliliter (g/mL) for liquids. In the context of this exercise, knowing the solution density allows us to find the volume that a certain mass occupies.

For the glucose solution with a given density of \(1.03 \text{g/mL}\), we use the formula: \
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Thus, a \(100 \text{g}\) mass of this solution would occupy \(97.087 \text{mL}\) which we convert to liters to use in molarity calculations, yielding \(0.097087 \text{L}\).

Understanding how to apply solution density is key in determining the volume and, subsequently, the concentration of a solution, demonstrated by the fact that this property is a bridge between mass and volume, both of which are essential variables in molarity calculations.