Question

A laboratory procedure calls for making 400.0 \(\mathrm{mL}\) of a 1.1 \(\mathrm{M} \mathrm{NaNO}_{3}\) solution. What mass of \(\mathrm{NaNO}_{3}(\mathrm{in} \mathrm{g})\) do you need?

Short answer

To prepare 400.0 mL of a 1.1 M \(\mathrm{NaNO}_{3}\) solution, 37.4 g of \(\mathrm{NaNO}_{3}\) is required.

Step-by-step solution

Determine the moles of solute needed

To find the mass of \(\mathrm{NaNO}_3\) required, we must first calculate the number of moles needed using the formula: molarity (M) = moles of solute / volume of solution (L). Rearranging the formula gives us: moles of solute = molarity × volume. For a 1.1 M solution, we need 1.1 moles of \(\mathrm{NaNO}_3\) per liter of solution. Since the desired volume is 400.0 mL or 0.400 L, the moles of solute needed are 1.1 moles/L × 0.400 L.

Calculate the mass of the solute

Calculate the molar mass of \(\mathrm{NaNO}_3\) using the atomic masses from the periodic table: \(\mathrm{Na} = 22.99 \(\mathrm{g/mol}\), \mathrm{N} = 14.01 \(\mathrm{g/mol}\), \mathrm{O} = 16.00 \(\mathrm{g/mol}\)\). The molar mass is the sum of the atomic masses: \(M_{\mathrm{NaNO}_3} = 22.99 \(\mathrm{g/mol}\) + 14.01 \(\mathrm{g/mol}\) + (3 \times 16.00 \(\mathrm{g/mol}\))\). The molar mass of \(\mathrm{NaNO}_3\) is then calculated to be approximately 85.00 g/mol. Next, multiply the moles of solute needed by the molar mass to find the mass in grams: mass = moles × molar mass.

Solve for the mass of \(\mathrm{NaNO}_3\)

Using the moles calculated in step 1 and the molar mass from step 2, the required mass of \(\mathrm{NaNO}_3\) is: mass = 0.440 moles × 85.00 \(\mathrm{g/mol}\). This will give us the mass in grams needed to prepare the solution.

Key concepts

Molarity Calculation

Understanding molarity is fundamental in chemistry when dealing with solutions. Molarity, denoted as 'M', is the concentration of a solute in a solution. It's defined as the number of moles of solute per liter of solution. The formula to calculate molarity is \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). This calculation is essential when preparing a specific volume of a solution with a desired concentration.

For instance, to prepare a 1.1 M \( \mathrm{NaNO}_{3} \) solution, you would first determine the volume in liters, then use the molarity formula to calculate the moles of solute needed. In our example, we have a volume of 0.400 liters and a target molarity of 1.1 M. Multiplying these gives us the moles of \( \mathrm{NaNO}_{3} \) required for the solution, which is \( 0.440 \) moles. Ensuring the accuracy of this step is crucial as it sets the foundation for the subsequent steps in the solution preparation.

Molar Mass

The molar mass is another crucial concept in chemistry, particularly in solution preparation. It represents the mass of one mole of a substance, measured in grams per mole (g/mol). It can be calculated by adding up the atomic masses of all the atoms in a compound, which are found on the periodic table.

To calculate the molar mass of \( \mathrm{NaNO}_{3} \) for example, we sum the atomic masses of sodium (Na), nitrogen (N), and oxygen (O). Breaking it down, sodium has an atomic mass of 22.99 g/mol, nitrogen has 14.01 g/mol, and oxygen has 16.00 g/mol. Since there are three oxygen atoms in the compound, we multiply the atomic mass of oxygen by three. Therefore, the molar mass of \( \mathrm{NaNO}_{3} \) is \( 22.99 \text{ g/mol} + 14.01 \text{ g/mol} + (3 \times 16.00 \text{ g/mol}) \) which approximately equals to 85.00 g/mol. This value is used to determine the mass of the solute needed to achieve the desired concentration in a given volume.

Solute Mass Calculation

Once the molarity and the molar mass are determined, the final step in the solution preparation process is to calculate the mass of the solute needed. This is done by multiplying the molar mass by the number of moles of solute.

Following our \( \mathrm{NaNO}_{3} \) example, we use the molar mass of 85.00 g/mol and the moles of solute (0.440 moles) calculated from the molarity. The equation \( \text{mass} = \text{moles} \times \text{molar mass} \) is applied, yielding \( 0.440 \text{ moles} \times 85.00 \text{ g/mol} \) which gives us the mass of \( \mathrm{NaNO}_{3} \) needed in grams. Knowing how to accurately calculate the solute mass ensures that the desired solution concentration is achieved, which is vital for experiments and various applications in chemistry.