Question

(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly 750 \(\mathrm{mL}\) of solution. (b) How many moles of KBr are present in 150 \(\mathrm{mL}\) of a 0.112\(M\) solution? (c) How many milliliters of 6.1\(M \mathrm{HCl}\) solution are needed to obtain 0.150 \(\mathrm{mol}\) of \(\mathrm{HCl}\) ?

Short answer

The molarity of the Na₂CrO₄ solution is 0.103 M, there are 0.0168 moles of KBr present in the 150 mL of 0.112 M solution, and approximately 24.59 mL of the 6.1 M HCl solution is needed to obtain 0.150 mol of HCl.

Step-by-step solution

a) Calculate the molarity of the Na₂CrO₄ solution

1. Calculate the molar mass of Na₂CrO₄. To calculate the molarity of the Na₂CrO₄ solution, we first need to determine the molar mass of Na₂CrO₄ using its atomic weights: Na = 22.990 g/mol, Cr = 51.996 g/mol, and O = 15.999 g/mol. Molar mass of Na₂CrO₄ = (2 × 22.990) + 51.996 + (4 × 15.999) = 45.98 + 51.996 + 63.996 = 161.972 g/mol 2. Calculate the moles of Na₂CrO₄. Now we can determine the moles of Na₂CrO₄ by dividing the mass of the solute (12.5 g) by its molar mass (161.972 g/mol). Moles of Na₂CrO₄ = 12.5 g / 161.972 g/mol = 0.07721 mol 3. Calculate the molarity of the Na₂CrO₄ solution. Molarity is defined as the moles of solute per liter of solution. The volume of the solution is given in mL, so first, we need to convert it to liters: 750 mL = 750 / 1000 = 0.750 L Now, we can calculate the molarity of the Na₂CrO₄ solution by dividing the moles of solute (0.07721 mol) by the volume of the solution (0.750 L). Molarity of Na₂CrO₄ solution = 0.07721 mol / 0.750 L = 0.103 M

b) Calculate the moles of KBr present in the solution

1. Convert the volume of the solution to liters. The volume of the KBr solution is initially given in mL. To find the moles of KBr present in the solution, we first need to convert the volume from mL to L: 150 mL = 150 / 1000 = 0.150 L 2. Calculate the moles of KBr. Now that we have the volume in liters, we can use the molarity of the KBr solution (0.112 M) to find the number of moles of KBr present in the solution: Moles of KBr = Molarity × Volume = 0.112 M × 0.150 L = 0.0168 mol

c) Calculate the volume of the 6.1 M HCl solution needed to obtain 0.150 mol of HCl

1. Calculate the volume of the 6.1 M HCl solution needed. To find the volume of the 6.1 M HCl solution needed to obtain 0.150 mol of HCl, we can use the formula: Volume = Moles of solute / Molarity Using the given molarity (6.1 M) and the moles of solute (0.150 mol), we can find the volume of the solution: Volume = 0.150 mol / 6.1 M = 0.02459 L 2. Convert the volume from liters to milliliters. Since the final answer is requested in milliliters, we can convert the volume from liters to milliliters: 0.02459 L × 1000 = 24.59 mL Approximately 24.59 mL of the 6.1 M HCl solution is needed to obtain 0.150 mol of HCl.

Key concepts

Moles

To understand chemistry calculations involving solutions, it's crucial to grasp the concept of moles. A mole is a standard scientific unit for measuring large quantities of very small entities such as atoms or molecules. One mole is equal to Avogadro's number, which is approximately 6.022 x 10²³ entities.

When you want to determine the amount of a substance in a chemical reaction, working with moles simplifies the math. Rather than dealing with tiny measurements like the weight of individual atoms, you're working with amounts that are more relatable in the laboratory setting.

In the exercise, to compute the moles of \( ext{Na}_2 ext{CrO}_4\) used, the formula used is:

• Moles = mass of the substance (g) / molar mass (g/mol)

This straightforward formula lets you convert the mass of a substance into moles, bridging the gap between the mass you can measure and the actual number of particles involved in the reaction.

Solution Concentration

Solution concentration indicates how much solute is present in a given quantity of solvent or solution. Molarity is a common measure of concentration, which represents the number of moles of solute per liter of solution. The formula for molarity (M) is shown as:

• Molarity (M) = Moles of solute / Volume of solution in liters

This unit is very handy in chemistry because it combines both the chemical nature of the solute (number of molecules) and the physical volume of the solvent.

In practical terms, knowing the molarity helps predict how a chemical reaction will proceed, as it informs you about the concentration of reactants in the solution. In the given exercise, calculating molarity helps to identify how concentrated the \( ext{Na}_2 ext{CrO}_4\) in the solution is, which allows for precise control over its use in further reactions.

Volume Conversion

Often in chemical calculations, volumes need to be converted from milliliters to liters to match the units used in molarity calculations.

A quick reminder is that 1 liter equals 1000 milliliters. Knowing this conversion is crucial because many laboratory measurements are made in milliliters, whereas molarity is expressed with liters. Therefore, you should always convert volumes to liters before plugging them into molarity or moles formulas.

In the exercise:

• 750 mL = 750 \( \div \) 1000 = 0.750 L
• 150 mL = 150 \( \div \) 1000 = 0.150 L

These conversions ensure consistency with units and are vital for accurate calculations. Failing to do so can lead to significant errors in determining the concentration or number of moles.

Molar Mass

Molar mass is a term used to describe the mass of one mole of a substance, expressed in grams per mole (g/mol). Each element has its molar mass, found on the periodic table, and the molar mass of a compound can be determined by summing the molar masses of its constituent atoms.

For example, in the case of \( ext{Na}_2 ext{CrO}_4\), its molar mass is calculated by adding up the molar masses of its elements:

• 2 sodium (Na) atoms: 2 \( \times \) 22.990 g/mol
• 1 chromium (Cr) atom: 51.996 g/mol
• 4 oxygen (O) atoms: 4 \( \times \) 15.999 g/mol

The sum gives the molar mass of the compound, 161.972 g/mol. Having the precise molar mass allows for exact conversions from mass to moles, which is critical for preparing solutions with accurate concentrations, such as in the given exercise.