The compound ammonia, NH3, is a weak electrolyte. Describe in at least two sentences how we know that ammonia is a weak electrolyte, and not a strong electrolyte or a nonelectrolyte.
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The compound ammonia, NH3, is a weak electrolyte. Describe in at least two sentences how we know that ammonia is a weak electrolyte, and not a strong electrolyte or a nonelectrolyte.
If a NaCl solution is 15.5% (m/m), how many grams of NaCl are in 250.0 grams of solution?
You are asked to dilute 10.0 mL of 12.0 M HCl with enough water to make a 1.09 solution. How much water is added for the dilution? Answer in mL to one decimal place.
If you were to place a red blood cell in a solution of water where it had no salt or glucose, what would happen to it? Please describe using the terms provided in the textbook regarding isotonic, hypertonic, and hypotonic solutions and red blood cells. You may use your textbook.
The correct answer and explanation is:
Here are detailed answers and explanations to your questions:
1. Ammonia as a Weak Electrolyte
Ammonia (NH₃) is classified as a weak electrolyte because it does not completely ionize in water. Instead, it reacts with water to form a small amount of ammonium ions (NH₄⁺) and hydroxide ions (OH⁻), leaving most of the ammonia molecules un-ionized in solution. This partial ionization leads to low electrical conductivity, distinguishing it from strong electrolytes that completely dissociate in water and nonelectrolytes that do not ionize at all.
2. NaCl in a 15.5% Solution
If a NaCl solution is 15.5% (m/m), this means that 15.5 grams of NaCl are present in every 100 grams of solution. To find how many grams of NaCl are in 250.0 grams of solution: Mass of NaCl=15.5 g100 g×250.0 g=38.75 g\text{Mass of NaCl} = \frac{\text{15.5 g}}{\text{100 g}} \times \text{250.0 g} = \text{38.75 g}
Answer: There are 38.75 grams of NaCl in 250.0 grams of the solution.
3. Dilution of HCl Solution
The dilution formula is: M1V1=M2V2M_1V_1 = M_2V_2
Where:
- M1M_1 = initial concentration = 12.0 M
- V1V_1 = initial volume = 10.0 mL
- M2M_2 = final concentration = 1.09 M
- V2V_2 = final volume (unknown)
Rearranging to find V2V_2: V2=M1×V1M2=12.0×10.01.09=110.1 mLV_2 = \frac{M_1 \times V_1}{M_2} = \frac{12.0 \times 10.0}{1.09} = 110.1 \, \text{mL}
The amount of water added is: Water added=V2−V1=110.1 mL−10.0 mL=100.1 mL\text{Water added} = V_2 – V_1 = 110.1 \, \text{mL} – 10.0 \, \text{mL} = 100.1 \, \text{mL}
Answer: 100.1 mL of water is added.
4. Red Blood Cell in Pure Water
When a red blood cell is placed in pure water, which is a hypotonic solution, water enters the cell through osmosis. The hypotonic solution has a lower solute concentration than the cytoplasm of the red blood cell. As water moves into the cell, the cell swells and may eventually burst, a process called hemolysis.
This occurs because water moves from an area of lower solute concentration (the pure water) to an area of higher solute concentration (inside the cell) to achieve equilibrium. In contrast, if the cell were in an isotonic solution, there would be no net water movement, maintaining cell shape. In a hypertonic solution, water would leave the cell, causing it to shrink or undergo crenation.
Explanation on Red Blood Cells in Solutions
Red blood cells are highly sensitive to the osmotic environment. When placed in a hypotonic solution such as pure water, the external solution has a lower solute concentration compared to the cell’s cytoplasm. Water moves into the cell via osmosis, driven by the gradient of water concentration. As water accumulates, the cell swells due to the increased internal pressure. If the influx continues unchecked, the cell membrane may rupture, resulting in hemolysis. This phenomenon demonstrates the principle of osmosis, where water moves toward a region of higher solute concentration to balance solute levels.
In an isotonic solution, the concentration of solutes inside the cell equals that of the external environment. This balance prevents net water movement, maintaining the red blood cell’s normal shape and function. Physiological saline (0.9% NaCl) is an example of an isotonic solution used in medical settings to avoid disrupting red blood cells.
Conversely, in a hypertonic solution, the solute concentration outside the cell is higher than inside. Water moves out of the cell, causing it to lose volume and shrink, a process known as crenation. This can impair the cell’s ability to function and may lead to cell death if the condition persists.
Understanding these processes is crucial in medical and biological contexts, as maintaining the correct osmotic balance is essential for cell survival and overall homeostasis.