Question

You make 1.000 L of an aqueous solution that contains \(35.0 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),(\mathbf{a})\) What is the molarity of sucrose in this solution? (b) How many liters of water would you have to add to this solution to reduce the molarity you calculated in part (a) by a factor of two?

Short answer

The molarity of sucrose is 0.1023 M. Add 1.000 L of water to reduce it by a factor of two.

Step-by-step solution

Calculate the molar mass of sucrose

The molecular formula of sucrose is \( \text{C}_{12} \text{H}_{22} \text{O}_{11} \). Calculate its molar mass by adding the molar masses of all atoms in the formula. The molar mass of carbon is \( 12.01 \, \text{g/mol} \), hydrogen is \( 1.01 \, \text{g/mol} \), and oxygen is \( 16.00 \, \text{g/mol} \). Thus, the molar mass of sucrose is: \[ 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.30 \, \text{g/mol} \].

Calculate the initial moles of sucrose

Use the mass of sucrose and its molar mass to calculate the moles of sucrose in the solution. The formula for calculating moles is \( n = \frac{\text{mass}}{\text{molar mass}} \). Therefore, \( n = \frac{35.0 \, \text{g}}{342.30 \, \text{g/mol}} \approx 0.1023 \, \text{mol} \).

Calculate the initial molarity of the solution

The molarity of a solution is calculated using \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume of the solution in liters. The volume of the solution is \( 1000 \, \text{L} \). So, \( M = \frac{0.1023 \, \text{mol}}{1.000 \, \text{L}} = 0.1023 \, \text{M} \).

Determine the new molarity and required volume

To reduce the molarity by a factor of two, the new molarity should be \( 0.1023 \, \text{M} / 2 = 0.0512 \, \text{M} \). The number of moles of sucrose remains the same (\( 0.1023 \, \text{mol} \)), so the new volume

Calculate the extra volume of water needed

Calculate the new volume needed for this molarity using \( M_{\text{new}} = \frac{n}{V_{\text{new}}} \), which gives \( V_{\text{new}} = \frac{0.1023 \, \text{mol}}{0.0512 \, \text{M}} \approx 2.000 \, \text{L} \). The original volume was \( 1.000 \, \text{L} \), so you need to add \( 2.000 \, \text{L} - 1.000 \, \text{L} = 1.000 \, \text{L} \) of water.

Key concepts

Molecular Formula

When working with solutions and compounds, understanding the molecular formula is key. The molecular formula reveals the specific numbers and types of atoms within a molecule. For example, sucrose has the molecular formula \( \text{C}_{12} \text{H}_{22} \text{O}_{11} \), indicating it is made up of 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms.

This formula is crucial because it helps determine other vital properties like molecular weight, which in turn is essential for calculating molar mass.

• Carbon contributes significantly to the molar mass due to its relative weight and number in sucrose.
• Hydrogen atoms, though smaller, add to the compound's structure and weight.
• Oxygen, a substantial part of the formula, affects the solubility and chemical reactions of sucrose.

Knowing the molecular weight helps in computations involving moles and molarity.

Moles of Sucrose

The concept of moles is fundamental in chemistry and is particularly important when dealing with solutions. A mole refers to Avogadro's number, approximately \(6.022 \times 10^{23}\) entities, which helps bridge the gap between atomic and macroscopic scales.

To calculate the number of moles in a given mass of sucrose, use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
For instance, with 35.0 g of sucrose and a molar mass of 342.30 g/mol, the moles of sucrose are calculated as follows:
\[ n = \frac{35.0}{342.30} \approx 0.1023 \text{ mol} \]

• Using moles makes stoichiometry simpler, connecting mass with a number of atoms.
• The mole concept is critical for understanding chemical reactions and proportions.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It provides a necessary link between mass and the number of particles.

The molar mass of sucrose \( (\text{C}_{12} \text{H}_{22} \text{O}_{11}) \) can be calculated by summing the atomic masses of all the atoms in the molecular formula. The atomic masses are as follows: carbon (12.01 g/mol), hydrogen (1.01 g/mol), and oxygen (16.00 g/mol). To find the molar mass, calculate:
\[ 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.30 \, \text{g/mol} \]

• The molar mass is vital for converting between mass and moles.
• This conversion is essential for chemical calculations and solution preparation.

Solution Dilution

Dilution is a process used to reduce the concentration of a solute in a solution. It's done by adding more solvent, keeping the number of moles of solute constant.

To dilute a solution's concentration by a factor, adjust the volume accordingly. In the given exercise, reducing the molarity of sucrose from 0.1023 M to 0.0512 M involves increasing the solution volume while maintaining the same number of moles of sucrose. The new total volume is found using the formula:
\[ M_{\text{new}} = \frac{n}{V_{\text{new}}} \]
Thus:
\[ V_{\text{new}} = \frac{0.1023}{0.0512} \approx 2.000 \, \text{L} \]
This means adding water to extend the total volume from 1.000 L to 2.000 L. Therefore, 1.000 L extra water is needed.

• Knowing how to dilute solutions is crucial in preparing solutions of desired concentration.
• Solution dilution principles are applicable in many practical and laboratory settings.