A biochemist carefully measures the molarity of salt in 84.1 mL of photobacterium cell growth medium to be 46.4 mM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 30.9 mL. Calculate the new molarity of salt in the photobacterium cell growth medium. Round each of your answers to 3 significant digits. ☐mM 0 x10 X S
A biochemist carefully measures the molarity of salt in 84.1 mL of photobacterium cell growth medium to be 46.4 mM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 30.9 mL. Calculate the new molarity of salt in the photobacterium cell growth medium. Round each of your answers to 3 significant digits. ☐mM 0 x10 X S
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter16: Solutions
Section: Chapter Questions
Problem 16.3TC
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![### Understanding Molarity Changes Due to Evaporation
**Problem Statement:**
A biochemist carefully measures the molarity of salt in 84.1 mL of photobacterium cell growth medium to be 46.4 mM.
Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 30.9 mL. Calculate the new molarity of salt in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.
**Solution:**
To solve for the new molarity after the evaporation of the solvent, we can use the formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) = initial concentration (46.4 mM)
- \( V_1 \) = initial volume (84.1 mL)
- \( C_2 \) = final concentration (unknown)
- \( V_2 \) = final volume (30.9 mL)
First, rearrange the formula to solve for the final concentration (\( C_2 \)):
\[ C_2 = \frac{C_1V_1}{V_2} \]
Plug in the known values:
\[ C_2 = \frac{46.4 \, \text{mM} \times 84.1 \, \text{mL}}{30.9 \, \text{mL}} \]
Calculate \( C_2 \):
\[ C_2 \approx \frac{3901.04}{30.9} \approx 126.2 \, \text{mM} \]
Therefore, the new molarity of the salt in the photobacterium cell growth medium after evaporation is approximately **126 mM** (rounded to three significant digits).
**Interactive Element Explanation:**
The interactive element provides a box where users can input the calculated molarity in mM (millimolar). There are two buttons available:
- A checkmark button (✓) to submit the answer.
- A refresh button to reset and recalculate if needed.
This problem is an excellent example of how changes in volume can affect the concentration of solutions, which is a fundamental concept in chemistry and biochemistry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2dc05801-0f2f-4b82-9bd0-eb38a3fb827a%2F80de3ec5-ad34-4fd5-b714-7c3b6f493b14%2Figc1n48_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Molarity Changes Due to Evaporation
**Problem Statement:**
A biochemist carefully measures the molarity of salt in 84.1 mL of photobacterium cell growth medium to be 46.4 mM.
Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 30.9 mL. Calculate the new molarity of salt in the photobacterium cell growth medium. Round each of your answers to 3 significant digits.
**Solution:**
To solve for the new molarity after the evaporation of the solvent, we can use the formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) = initial concentration (46.4 mM)
- \( V_1 \) = initial volume (84.1 mL)
- \( C_2 \) = final concentration (unknown)
- \( V_2 \) = final volume (30.9 mL)
First, rearrange the formula to solve for the final concentration (\( C_2 \)):
\[ C_2 = \frac{C_1V_1}{V_2} \]
Plug in the known values:
\[ C_2 = \frac{46.4 \, \text{mM} \times 84.1 \, \text{mL}}{30.9 \, \text{mL}} \]
Calculate \( C_2 \):
\[ C_2 \approx \frac{3901.04}{30.9} \approx 126.2 \, \text{mM} \]
Therefore, the new molarity of the salt in the photobacterium cell growth medium after evaporation is approximately **126 mM** (rounded to three significant digits).
**Interactive Element Explanation:**
The interactive element provides a box where users can input the calculated molarity in mM (millimolar). There are two buttons available:
- A checkmark button (✓) to submit the answer.
- A refresh button to reset and recalculate if needed.
This problem is an excellent example of how changes in volume can affect the concentration of solutions, which is a fundamental concept in chemistry and biochemistry.
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