Problem 85 - Chemistry and Chemical Reactivity

Question

Which contains the greater mass of solute: 1 I. of \(0.1 \mathrm{M}\) NaCl or 1 L of \(0.06 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Step-by-Step Solution

Answer

1 L of 0.06 M Na2CO3 contains the greater mass of solute.

Step 1: Identify the Formula for Molarity

Molarity (\(M\)) is given by the number of moles of solute per liter of solution. Mathematically, it is\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]We know the molarity and volume of each solution.

Step 2: Calculate Moles of NaCl

For \(1 ext{ L of } 0.1 ext{ M } ext{NaCl}\), the number of moles can be calculated using\(moles = M imes ext{volume} = 0.1 ext{ moles/L} imes 1 ext{ L} = 0.1 ext{ moles}\).

Step 3: Calculate Moles of Na2CO3

For\(1 ext{ L of } 0.06 ext{ M } ext{Na}_{2} ext{CO}_{3}\), calculate the moles:\(moles = 0.06 ext{ moles/L} imes 1 ext{ L} = 0.06 ext{ moles}\).

Step 4: Find Molar Masses

Calculate the molar masses of NaCl and\( ext{Na}_{2} ext{CO}_{3}\):- \( ext{NaCl}\): \(Na = 23 ext{ g/mol}\), \(Cl = 35.5 ext{ g/mol}\) adding to \(58.5 ext{ g/mol}\).- \( ext{Na}_{2} ext{CO}_{3}\): \(2 imes 23 ext{ g/mol (Na)} + 12 ext{ g/mol (C)} + 3 imes 16 ext{ g/mol (O)} = 106 ext{ g/mol}\).

Step 5: Calculate Mass of NaCl

Multiply the moles of\( ext{NaCl}\) by its molar mass:\[0.1 ext{ moles} imes 58.5 ext{ g/mol} = 5.85 ext{ g}\].

Step 6: Calculate Mass of Na2CO3

Multiply the moles of\( ext{Na}_{2} ext{CO}_{3}\) by its molar mass:\[0.06 ext{ moles} imes 106 ext{ g/mol} = 6.36 ext{ g}\].

Step 7: Compare Masses

Compare the calculated masses: \(5.85 ext{ g (NaCl)}\) vs. \(6.36 ext{ g (Na}_{2} ext{CO}_{3}\) to determine which has the greater mass of solute. \( ext{Na}_{2} ext{CO}_{3}\) has the greater mass.

Key Concepts

Solute Mass

When we talk about the solute mass, we're referring to the mass of the substance that is being mixed into a solution. It's essential in chemistry to determine which solution has a larger amount of solute, as it can affect the properties and behaviors of the solution.
To calculate the solute mass, you first need to know two things:

• The number of moles of the solute in the solution.
• The molar mass of the solute.

For example, in a solution of NaCl, once you figure out how many moles of NaCl are present, you multiply it by the molar mass of NaCl to get the total solute mass. This step makes it easier to compare with other solutes, such as Na\(_2\)CO\(_3\), to determine which has a greater solute mass.

Moles Calculation

Calculating moles is a fundamental step in finding the amount of solute. The number of moles is essentially a way to quantify the amount of substance using Avogadro's number.
To calculate the moles present in a solution, you use the molarity formula:

• Molarity (\(M\)) = moles of solute per liter of solution.
• Thus, moles = Molarity \(\times\) Volume of the solution in liters.

In the exercise, for NaCl (at 0.1 M), multiplying the molarity by the volume (1 L) gives 0.1 moles. Similarly, for Na\(_2\)CO\(_3\) (at 0.06 M), the calculation results in 0.06 moles. This process provides the groundwork for further calculations, such as finding the solute mass.

Molar Mass

Understanding molar mass is crucial for converting between moles and grams, which are often more practical to measure. Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance (mole). It is usually expressed in grams per mole (g/mol).
The molar mass is calculated by summing the atomic masses of all atoms in the formula of the compound.

• For NaCl, you add the mass of Na (23 g/mol) and Cl (35.5 g/mol), resulting in 58.5 g/mol.
• For Na\(_2\)CO\(_3\), you multiply and add the molar masses of its individual atoms: 2 Na atoms (2 \( \times \) 23 g/mol), 1 C atom (12 g/mol), and 3 O atoms (3 \( \times \) 16 g/mol), totaling 106 g/mol.

With molar mass, you can convert the moles of a substance into grams, helping you find the mass of the solute in your solution. This knowledge is indispensable for comparing solutes or conducting experiments that require precise measurements.