How would you prepare 1.00 L of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 \mathrm{M})\) sulfuric acid b. HCl from "concentrated" (12 M) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO_ from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

Molarity is a measurement of concentration, representing the number of moles of a solute per liter of solution. Knowing how to calculate molarity is essential in chemistry, especially when preparing solutions.

Molarity, denoted by the symbol 'M', is calculated using the formula: \[ M = \frac{{n}}{{V}} \]where 'n' is the number of moles of the solute and 'V' is the volume of the solution in liters. To find the molarity of a solution, we can rearrange this formula to solve for the unknown quantity, as long as we have the other two.

In the practice of preparing solutions, a common task is to dilute a concentrated solution to attain a desired molarity. The dilution equation, \[ M_1V_1 = M_2V_2 \], is vital. This equation means that the product of the original concentration (\( M_1 \)) and the original volume (\( V_1 \)) of the solution equals the product of the final concentration (\( M_2 \)) and the final volume (\( V_2 \)). It's widely used to calculate how much of the concentrated stock solution is needed to achieve a specific concentration and volume of a new solution.

When working with chemical solutions, sometimes we need to lower the concentration by adding more solvent, a process known as dilution. The concept of dilution is fundamental in labs for achieving desired solution concentrations without having to prepare a fresh batch every time.

To dilute a solution, one simply takes a measure of the stock solution and adds solvent (commonly water) until reaching the desired concentration. The molarity equation \( M_1V_1 = M_2V_2 \) provides a straightforward method to determine how much of the stock solution is needed for a given dilution. In essence, you are keeping the amount of substance constant (moles), while increasing the volume.

It is important to acknowledge that while diluting, the 'solute' remains the same, only the volume of 'solvent' changes. Thus, understanding dilution enables one to prepare solutions of accurate molarity efficiently, which is crucial for experimental consistencies and cost-effectiveness in a laboratory setting.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a physical property that is used for many calculations in chemistry, such as when converting between mass and moles of a substance.

To determine the molar mass, one needs to sum the atomic masses of all the atoms in the compound. Each element's atomic mass is found on the periodic table, and for a molecule, the total molar mass is the sum of the molar masses of each individual element multiplied by the number of atoms of that element in the molecule. For hydrates, water molecules associated with the compound must also be included in the calculation.

For example, in the exercise, the molar mass of \( NiCl_2 \bullet 6H_2O \) was calculated by adding the molar masses of nickel (Ni), chlorine (Cl), hydrogen (H), and oxygen (O) corresponding to their molecular formula. The calculation included the atoms from the six water molecules (\( 6H_2O \)) that are part of the crystal structure. Determining the correct molar mass is vital as it directly impacts the preparation of solutions with accurate molarity.