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Chapter 12: Problem 103

Your assistant tells you she measured out \(2.50\) moles of \(\mathrm{NaCl}\) and then added enough water to get \(500.0 \mathrm{~mL}\) of solution to prepare a \(5.00 \mathrm{M}\) solution of \(\mathrm{NaCl}\). (a) What was the mass of the \(\mathrm{NaCl}\) in grams? (b) Did she successfully prepare a \(2.5 \mathrm{M}\) solution? Prove your answer.

Short Answer

Expert verified
(a) The mass of the NaCl in grams is calculated as: Mass (g) = Moles × Molar mass = 2.50 moles × 58.44 g/mol = 146.1 g (b) No, the student did not successfully prepare a 2.5 M solution. The observed molarity of the solution is: Molarity (M) = 2.50 moles / 0.5 L = 5.00 M Since the observed molarity (5.00 M) is not equal to the target molarity (2.5 M), the student did not prepare a 2.5 M solution.

Step by step solution

01

Find the mass of NaCl

We are given the amount of NaCl in moles which is 2.50 moles. We need to convert it to mass (grams). To do this, we'll use the molar mass of NaCl which is approximately 58.44 g/mol. The formula to find the mass is: Mass (g) = Moles × Molar mass Now, plug the values into the formula: Mass (g) = 2.50 moles × 58.44 g/mol
02

Calculate the mass of NaCl

Now, we'll calculate the mass of NaCl: Mass (g) = 2.50 moles × 58.44 g/mol = 146.1 g So, 146.1 grams of NaCl were used in the solution preparation.
03

Find the molarity of the prepared solution

The student aimed to prepare a 5.00 M solution, but we will verify if she actually prepared a 2.5 M solution or not. Molarity is defined by the formula: Molarity (M) = Moles of solute / Volume of solution (L) We know the number of moles of NaCl (2.5 moles) and the final volume of the solution (500.0 mL). To find the molarity, we need to convert the volume from milliliters to liters: Volume (L) = 500.0 mL × (1 L / 1000 mL) = 0.5 L Now, we calculate the molarity: Molarity (M) = 2.50 moles / 0.5 L
04

Calculate the molarity of the prepared solution

Now, we'll calculate the observed molarity of the prepared solution: Molarity (M) = 2.50 moles / 0.5 L = 5.00 M The observed molarity of the prepared solution is 5.00 M.
05

Check if the 2.5 M solution was prepared successfully

Now, we will compare the observed molarity with the target molarity (2.5 M) to answer part (b) of the exercise: Observed molarity: 5.00 M (calculated in step 4) Target molarity: 2.5 M (mentioned in the question) Since the observed molarity (5.00 M) is not equal to the target molarity (2.5 M), the student did not prepare a 2.5 M solution. So the answer to part (a) is 146.1 grams, and the answer to part (b) is no, the student did not successfully prepare a 2.5 M solution.

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