To determine the molality of a [tex]$D$[/tex]-glucose solution prepared by dissolving [tex]\(54.0\)[/tex] grams of [tex]\(D\)[/tex]-glucose ([tex]\(C_6H_{12}O_6\)[/tex]) in [tex]\(125.9\)[/tex] grams of water, follow these detailed steps:
1. Determine the molar mass of [tex]\(D\)[/tex]-glucose ([tex]\(C_6H_{12}O_6\)[/tex]):
The molar mass is given as [tex]\(180.16\)[/tex] g/mol.
2. Calculate the number of moles of [tex]\(D\)[/tex]-glucose:
Use the formula:
[tex]\[
\text{moles of \(D\)-glucose} = \frac{\text{mass of \(D\)-glucose}}{\text{molar mass of \(D\)-glucose}}
\][/tex]
Given the mass of [tex]\(D\)[/tex]-glucose ([tex]\(54.0\)[/tex] grams):
[tex]\[
\text{moles of \(D\)-glucose} = \frac{54.0 \, \text{g}}{180.16 \, \text{g/mol}} \approx 0.2997 \, \text{mol}
\][/tex]
3. Convert the mass of the solvent (water) to kilograms:
Given the mass of the solvent ([tex]\(125.9\)[/tex] grams):
[tex]\[
\text{solvent mass} = \frac{125.9 \, \text{g}}{1000} = 0.1259 \, \text{kg}
\][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]) of the solution:
Molality is defined as the number of moles of solute per kilogram of solvent. Use the formula:
[tex]\[
\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
\][/tex]
Substitute the calculated values:
[tex]\[
\text{molality} = \frac{0.2997 \, \text{mol}}{0.1259 \, \text{kg}} \approx 2.38 \, m
\][/tex]
Therefore, the molality of the [tex]\(D\)[/tex]-glucose solution is [tex]\(2.38 \, m\)[/tex]. The correct answer is:
[tex]\[
\boxed{2.38 \, m}
\][/tex]
