Problem 68 - Chemistry: The Central Science

Question

(a) How many grams of solute are present in \(50.0 \mathrm{~mL}\) of \(0.488 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (b) If \(4.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in enough water to form \(400 \mathrm{~mL}\) of solution, what is the molarity of the solution? (c) How many milliliters of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) contain \(1.75 \mathrm{~g}\) of solute?

Step-by-Step Solution

Answer

(a) 7.18 grams of K2Cr2O7 are present in the 50.0 mL solution. (b) The molarity of the (NH4)2SO4 solution is 0.0758 M. (c) 438.8 mL of 0.0250 M CuSO4 solution contain 1.75 g of solute.

Step 1: Part (a): Finding grams of solute in a solution

1. Use the formula for molarity: \( M = \frac{n}{V} \) where \(M\) is molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters. We want to find the moles of K2Cr2O7 in the solution, so we can rewrite the formula for n: \(n = M \times V\) 2. Convert the volume of the solution to liters: 50.0 mL × \( \frac{1 L}{1000 mL} \) = 0.0500 L 3. Calculate the number of moles of K2Cr2O7 in the solution: \(n = M \times V \) = 0.488M × 0.0500L = 0.0244 moles 4. Calculate the molar mass of K2Cr2O7 (using atomic masses: K = 39.10 g/mol, Cr = 51.996 g/mol, O = 16.00 g/mol): Molar mass = (2 × 39.10) + (2 × 51.996) + (7 × 16.00) = 294.2 g/mol 5. Calculate the grams of K2Cr2O7 in the solution: \(mass = moles \times molar\,mass \) = 0.0244 moles × 294.2 g/mol = 7.18g The solution contains 7.18 grams of K2Cr2O7.

Step 2: Part (b): Finding the molarity of a solution

1. Calculate the molar mass of (NH4)2SO4 (using atomic masses: N = 14.01 g/mol, H = 1.008 g/mol, S = 32.07 g/mol, O = 16.00 g/mol): Molar mass = (2 × ((1 × 14.01) + (4 × 1.008))) + (1 × 32.07) + (4 × 16.00) = 132.14 g/mol 2. Calculate the number of moles of (NH4)2SO4 dissolved in the solution: \(moles = \frac{mass}{molar\,mass} \) = \( \frac{4.00 g}{132.14 g/mol} \) = 0.0303 moles 3. Convert the volume of the solution to liters: 400 mL × \( \frac{1 L}{1000 mL} \) = 0.400 L 4. Calculate the molarity of the solution: \( M = \frac{n}{V} \) = \( \frac{0.0303\,moles}{0.400 L} \) = 0.0758 M The molarity of the (NH4)2SO4 solution is 0.0758 M.

Step 3: Part (c): Finding the volume of a solution containing a specific amount of solute

1. Calculate the molar mass of CuSO4 (using atomic masses: Cu = 63.55 g/mol, S = 32.07 g/mol, O = 16.00 g/mol): Molar mass = (1 × 63.55) + (1 × 32.07) + (4 × 16.00) = 159.62 g/mol 2. Find the number of moles of CuSO4 in 1.75 g of solute: \(moles = \frac{mass}{molar\,mass} \) = \( \frac{1.75 g}{159.62 g/mol} \) = 0.01097 moles 3. Use the molarity formula to find the volume of the solution: \( V = \frac{n}{M} \) = \( \frac{0.01097\,moles}{0.0250 M} \) = 0.4388 L 4. Convert the volume to milliliters: 0.4388 L × \( \frac{1000 mL}{1 L} \) = 438.8 mL The volume of the 0.0250 M CuSO4 solution containing 1.75 g of solute is 438.8 mL.

Key Concepts

Molarity Calculations

Molarity is a crucial concept in chemistry that refers to the concentration of a solution. To calculate molarity, you need to know two key components: the number of moles of solute and the volume of the solution in liters. The formula used is:

• \[ M = \frac{n}{V} \]where \( M \) is molarity, \( n \) is the number of moles, and \( V \) is the volume in liters.

In the exercise, to find the solute's mass in solution, we first rearrange this formula to find the number of moles:

• \( n = M \times V \)

Then, using the formula \( \text{mass} = \text{moles} \times \text{molar mass} \), you can determine the solute's mass. Remember to always convert the volume from milliliters to liters by dividing by 1000, ensuring accuracy in calculations. These steps simplify the process of working out how much of a particular solute is found in a given solution.

Mole Concept

The mole concept is fundamental in chemistry, allowing chemists to count particles by weighing them. A mole is defined as Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) particles. This concept allows for easy conversion between the mass of a substance and the number of atoms or molecules it contains.
Understanding how to convert mass into moles is essential and involves using the formula:

• \[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

In part (b) of the example exercise, you were given a mass of \((NH_4)_2SO_4\) and needed to find the moles. Using the molar mass helps achieve this conversion, highlighting the connection between weight and the number of particles.
This principle aids in converting from one type of unit to another, making it easier to handle the substances involved in chemical reactions and solutions.

Molar Mass Determination

Determining molar mass is a small but vital step in solving chemistry problems. The molar mass is the mass of one mole of a compound and is expressed in grams per mole (g/mol). It combines the atomic masses of all atoms within a chemical formula.
For instance, to find the molar mass of \( K_2Cr_2O_7 \), you add:

• K: 2 atoms \(\times 39.10\) g/mol
• Cr: 2 atoms \(\times 51.996\) g/mol
• O: 7 atoms \(\times 16.00\) g/mol

Adding these values results in a molar mass of 294.2 g/mol, a necessary step used to find out how much of a solute's mass is present in a solution as in part (a) of the exercise.
Using the correct molar mass helps ensure precise calculations in converting between grams, moles, and ultimately determining the concentration or necessary volume of solutions in various chemical contexts.