Question

Determine the number of grams of solute needed to make each of the following molal solutions: a. a 4.50 \(\mathrm{m}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in 1.00 \(\mathrm{kg} \mathrm{H}_{2} \mathrm{O}\) b. a 1.00 \(\mathrm{m}\) solution of \(\mathrm{HNO}_{3}\) in 2.00 \(\mathrm{kg} \mathrm{H}_{2} \mathrm{O}\)

Short answer

441.36 grams of \(H_{2}SO_{4}\) and 126.02 grams of \(HNO_{3}\).

Step-by-step solution

Understanding Molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is: \[ m = \frac{n_{solute}}{m_{solvent} \text{ in kg}} \] where \( n_{solute} \) is the number of moles of solute.

Calculating Moles of Solute

For the given molality and mass of water, calculate the moles of solute using the formula: \[ n_{solute} = m \times m_{solvent} \]

Part a: Calculating moles of \( H_{2}SO_{4} \)

Given a 4.50 m solution of \( H_{2}SO_{4} \) in 1.00 kg of water, calculate the moles: \[ n_{H_{2}SO_{4}} = 4.50 \text{ m} \times 1.00 \text{ kg} \] \[ n_{H_{2}SO_{4}} = 4.50 \text{ moles} \]

Converting Moles to Grams

To find the mass of the solute, convert moles to grams using the molar mass. The molar mass of \( H_{2}SO_{4} \) is approximately 98.08 g/mol. Use the formula: \[ \text{mass} = n_{solute} \times \text{molar mass} \]

Part a: Converting moles of \( H_{2}SO_{4} \) to grams

Using the calculated moles: \[ \text{mass}_{H_{2}SO_{4}} = 4.50 \text{ moles} \times 98.08 \text{ g/mol} \] \[ \text{mass}_{H_{2}SO_{4}} = 441.36 \text{ grams} \]

Part b: Calculating moles of \( HNO_{3} \)

Given a 1.00 m solution of \( HNO_{3} \) in 2.00 kg of water, calculate the moles: \[ n_{HNO_{3}} = 1.00 \text{ m} \times 2.00 \text{ kg} \] \[ n_{HNO_{3}} = 2.00 \text{ moles} \]

Part b: Converting moles of \( HNO_{3} \) to grams

The molar mass of \( HNO_{3} \) is approximately 63.01 g/mol. Using the calculated moles: \[ \text{mass}_{HNO_{3}} = 2.00 \text{ moles} \times 63.01 \text{ g/mol} \] \[ \text{mass}_{HNO_{3}} = 126.02 \text{ grams} \]

Key concepts

Molality

Molality is an important concept in chemistry used to express the concentration of a solution. Unlike molarity, which depends on the volume of the solution, molality is defined as the number of moles of solute per kilogram of solvent. It is represented by the symbol \( \text{m} \). Calculating molality is straightforward when you have the necessary information.
The formula for molality is:\( m = \frac{n_{solute}}{m_{solvent} \text{ in kg}}\)
Here, \( n_{solute} \) is the number of moles of the solute, and \( m_{solvent} \) is the mass of the solvent in kilograms. Since molality is based on mass, it remains constant regardless of changes in temperature or pressure.
To understand this better, consider a scenario where we need to create a molal solution. Knowing the molality and the mass of the solvent can help us calculate the moles of solute needed. This is essential for various applications in chemistry, from industrial processes to academic experiments.

Moles of Solute

Understanding moles is crucial for making molal solutions. A mole is a unit representing \(6.022 \times 10^{23}\) entities (atoms, molecules, ions, etc.). This number is known as Avogadro's number. To calculate the number of moles of a solute in a solution, you use the given mass of solvent and the molality.
The formula is straightforward:\( n_{solute} = m \times m_{solvent} \)
For example, if you have a 4.50 \( \text{m} \) solution of \( \text{H}_{2}\text{SO}_{4} \) with 1.00 kg of solvent (water), it translates to: \( n_{solute} = 4.50 \text{ m} \times 1.00 \text{ kg} = 4.50 \text{ moles}\)
This means there are 4.50 moles of \( \text{H}_{2} \text{SO}_{4} \) in the solution. Understanding this step is fundamental as it forms the basis for further calculations, like converting these moles into grams to measure the physical amount of solute needed.

Molar Mass

Knowing the molar mass of a substance is essential when converting between moles and grams. The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It can be calculated by summing the atomic masses (from the periodic table) of all atoms in a molecule.
For example, the molar mass of \( \text{H}_{2} \text{SO}_{4} \) (sulfuric acid) involves two hydrogens (2 x 1.01 g/mol), one sulfur (32.07 g/mol), and four oxygens (4 x 16.00 g/mol):
\( 2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.08 \text{ g/mol}\)
Similarly, the molar mass of \( \text{HNO}_{3} \) is calculated as:
\( 1 \times 1.01 + 14.01 + 3 \times 16.00 = 63.01 \text{ g/mol}\)
Having these values allows us to convert moles of any compound into grams accurately, playing a critical role in preparing specific concentrations of solutions.

Grams of Solute

Once you have the number of moles and the molar mass of the solute, the next step is calculating the grams of solute needed. This conversion uses the formula:
\( \text{mass} = n_{solute} \times \text{molar mass} \)
For instance, using our earlier example, with 4.50 moles of \( \text{H}_{2} \text{SO}_{4} \) and a molar mass of 98.08 g/mol:
\( \text{mass}_{H_{2}\text{SO}_{4}} = 4.50 \text{ moles} \times 98.08 \text{ g/mol} = 441.36 \text{ grams}\)
Similarly, if you need grams of \( \text{HNO}_{3} \) and you have 2.00 moles, with a molar mass of 63.01 g/mol:
\( \text{mass}_{HNO_{3}} = 2.00 \text{ moles} \times 63.01 \text{ g/mol} = 126.02 \text{ grams}\)
These conversions are handy to directly measure the quantity of a solute to mix into a solvent for a desired solution concentration.

Solution Concentration

The concentration of a solution is a measure of the amount of solute dissolved in a given quantity of solvent or solution. In the context of molality, it specifically refers to the number of moles of solute per kilogram of solvent.
This form of concentration is helpful when dealing with situations where temperature changes are involved, as molality does not depend on volume changes that temperature can cause. To achieve a particular concentration:

• First determine the required molality and the mass of your solvent.
• Use the formula \( n_{solute} = m \times m_{solvent} \) to find out the amount of moles needed.
• Then convert those moles into grams using the molar mass of the solute.

For example, preparing a 1.00 \( \text{m} \) solution of \( \text{HNO}_{3} \) in 2.00 kg of water:

• Calculate moles: \( n_{HNO_{3}} = 1.00 \text{ m} \times 2.00 \text{ kg} = 2.00 \text{ moles}\)
• Convert to grams using molar mass: \( \text{mass}_{HNO_{3}} = 2.00 \text{ moles} \times 63.01 \text{ g/mol} = 126.02 \text{ grams}\)

Hence, ensuring you measure 126.02 grams of \( \text{HNO}_{3} \) and dissolve it in 2.00 kg of water will give you the desired 1.00 molal concentration.