Problem 44 - Chemistry The Central Science

Question

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) in \(175 \mathrm{~mL}\) of solution, (c) \(35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to \(0.500 \mathrm{~L} ?\)

Step-by-Step Solution

Answer

The molarities of the given solutions are: (a) 175.2 M, (b) 0.1195 M, and (c) 0.630 M.

Step 1: Convert mass of solute to moles

To calculate the moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), we have to convert the given mass of solute (15.0 g) into moles. We do this by using the formula: moles = \( \cfrac{mass \ of \ solute}{molar \ mass \ of \ solute} \) The molar mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is \(2(27.0) + 3(32.1+4(16.00))= 342.2 \mathrm{~g/mol}\). So, moles \(= \cfrac{15.0 \mathrm{~g}}{342.2 \mathrm{~g/mol}} = 0.04381 \mathrm{~mol}\) #a) Calculate volume of solution in liters#

Step 2: Convert volume in mL to L

Given the volume of solution in mL, we convert it into liters: Volume in L = \(0.250 \mathrm{~mL} \times \cfrac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.000250 \mathrm{~L}\) #a) Calculate molarity on the solution#

Step 3: Use the molarity formula

Now we can use the molarity formula for the solution: Molarity = \(\cfrac{0.04381 \mathrm{~mol}}{0.000250 \mathrm{~L}} = 175.2 \mathrm{~M}\) Molarity of solution (a) is 175.2 M. #b) Calculate moles of Mn(NO3)2•2H2O solute#

Step 4: Convert mass of solute to moles

To calculate the moles of \(\mathrm{Mn(NO}_{3}\mathrm{)_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}\), we have to convert the given mass of solute (5.25 g) into moles. The molar mass of \(\mathrm{Mn(NO}_{3}\mathrm{)_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O}\) is \((54.94+(14.01+3(16.00)) \times 2+4(1.008)+2(16.00)) = 251.0 g/mol\). So, moles \(= \cfrac{5.25 \mathrm{~g}}{251.0 \mathrm{~g/mol}} = 0.02092 \mathrm{~mol}\) #b) Calculate volume of solution in liters#

Step 5: Convert volume in mL to L

Given the volume of solution in mL, we convert it into liters: Volume in L = \(175 \mathrm{~mL} \times \cfrac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.175 \mathrm{~L}\) #b) Calculate molarity of the solution#

Step 6: Use the molarity formula

Now we can use the molarity formula for the solution: Molarity = \(\cfrac{0.02092 \mathrm{~mol}}{0.175 \mathrm{~L}} = 0.1195 \mathrm{~M}\) Molarity of solution (b) is 0.1195 M. #c) Determine amount of H2SO4 in initial solution#

Step 7: Calculate moles of H2SO4 from initial molarity and volume

Molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is given as 9.00 M and initial volume is 35.0 mL. We use the following formula to determine the moles of solute in the initial solution: Moles of solute = molarity × volume in L Moles of solute = \(9.00 \mathrm{~M} \times (35.0 \mathrm{~mL} \times \cfrac{1 \mathrm{~L}}{1000 \mathrm{~mL}}) = 0.315 \mathrm{~mol}\) #c) Calculate molarity of H2SO4 after dilution#

Step 8: Use the molarity formula

After diluting the solution to 0.500 L, we can now use the molarity formula to find the molarity: Molarity = \(\cfrac{0.315 \mathrm{~mol}}{0.500 \mathrm{~L}} = 0.630 \mathrm{~M}\) Molarity of solution (c) is 0.630 M.

Key Concepts

Moles of solute

Understanding moles is crucial for solving problems in chemistry, like those involving molarity. Moles are a way to count atoms or molecules, making it easy to measure the amount of a substance. To find the moles of a solute, use the formula:

• Moles = \( \frac{\text{mass of solute}}{\text{molar mass of solute}} \)

In the problems given, you need to determine the moles of the substances \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) and \( \mathrm{Mn(NO}_{3}\mathrm{)_{2}\cdot 2 \mathrm{H}_{2} \mathrm{O} \).
The key is knowing the molar mass of each compound. Molar mass is the mass of one mole of a substance and is calculated by adding the masses of each atom in a compound.
By converting the given mass of the solute to moles, you can engage in calculations involving concentrations and reactions.

Molar mass

Molar mass plays an essential role in converting grams to moles. It's like a bridge between the mass of a solute and the amount in moles. To find the molar mass of a compound, add together the atomic masses of all atoms in the chemical formula. For instance, in \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), calculate the molar mass by considering:

• 2 Al atoms (each 27.0 g/mol)
• 3 S atoms (each 32.1 g/mol)
• 12 O atoms (each 16.0 g/mol)

The total molar mass will help you convert grams to moles.
Similar steps apply to \(\mathrm{Mn(NO}_{3}\mathrm{)_{2}\cdot 2\mathrm{H}_{2} \mathrm{O} \) where calculating the molar mass involves considering Mn, N, O, and H atoms.
Using molar mass accurately is vital for any stoichiometric calculations.

Volume conversion

Volume conversion is an important step in calculating molarity. Always convert volumes to liters, as molarity is expressed in moles per liter. It’s simple to convert milliliters to liters:

• 1 mL = 0.001 L

In our problems:
For example, \(0.250 \mathrm{~mL} \) must be converted to \(0.000250 \mathrm{~L} \).
This change in volume makes the later molarity calculations precise.
Remember to double-check your unit conversions. Mistakes in volume conversion can skew results, leading to errors in understanding molarity.

Dilution

Dilution is a helpful technique when you need to decrease the concentration of a solution. It's based on the principle that the number of moles of solute remains constant before and after dilution. The equation for dilution is:

• \( \text{C}_{1} \times \text{V}_{1} = \text{C}_{2} \times \text{V}_{2} \)

Here, \(\text{C}_{1}\) and \(\text{V}_{1}\) are the initial concentration and volume, while \(\text{C}_{2}\) and \(\text{V}_{2}\) are the final concentration and volume.
In our example, with \( \mathrm{H}_{2} \mathrm{SO}_{4}\), start with a 9.00 M solution and use the volume change to find the new concentration after dilution.
This technique is vital in labs where specific concentrations are needed for experiments.