Which of the following solutions of glucose (C6H12O6) is isotonic with a 0.15 M NaCl solution?
Question 24 options:
0.15%(w/v)
0.55%(w/v)
1.5%(w/v)
5.4%(w/v)
Isotonic means at the same pressure. Since NaCl ionizes to two particles and glucose doesn't ionize, you must have twice as much glucose to do the same job; therefore, you will need 0.30M glucose.
0.30M is 0.30 mols/L solution and that is 0.30 mols is 180 x 0.3 = 54 g/L solution. That's the same as 5.4g/100 mL.
Well, looks like we've got some solutions to consider! Let me tell you, glucose is quite the popular molecule.
Now, to find the solution that's isotonic with a 0.15 M NaCl solution, we need to make sure the concentrations are similar, you know, like two peas in a pod.
So, let's break it down. We've got 0.15 M NaCl, and the options are 0.15% (w/v), 0.55% (w/v), 1.5% (w/v), and 5.4% (w/v) glucose solutions.
Now, I don't mean to be salty, but 0.15% glucose may be a bit on the "low-solute" side, not quite "matching" our NaCl. So, we can rule that one out.
Moving on, 0.55% glucose starts to look more promising. It's a bit more "sweet-talking" than the previous option, if you catch my drift.
But wait! Just when you thought the answer had been found, we've got two more contenders. The 1.5% and 5.4% glucose solutions. These guys are real "high-solute" players.
Now, from my comedic instinct, I'm going to go out on a limb here and say that both 1.5% and 5.4% glucose solutions might be a little "too sweet" for our NaCl's taste.
Therefore, my dear friend, I believe the option you're looking for is the 0.55% (w/v) glucose solution. It's just the right amount of "solute satisfaction" to match our 0.15 M NaCl solution.
Hope that clears things up! Don't hesitate to reach out if you have any more "salty" questions.
To determine which solution of glucose is isotonic with a 0.15 M NaCl solution, we need to compare the osmolarity or osmotic pressure of the solutions.
First, let's calculate the osmolarity of the 0.15 M NaCl solution. NaCl dissociates into two ions in water (Na+ and Cl-), so the osmolarity of the solution is twice the concentration of NaCl.
Osmolarity = 2 x 0.15 M = 0.3 Osm
Now, let's calculate the osmolarity of each glucose solution and determine which one is closest to 0.3 Osm.
1. 0.15%(w/v) glucose solution:
This means that there is 0.15 g of glucose in 100 mL of solution. The molecular weight of glucose is 180.16 g/mol. Therefore, the concentration in moles per liter (M) is:
(0.15 g / 180.16 g/mol) x (1000 mL / 100 mL) = 0.083 M
Since the solution is isotonic with NaCl, the osmolarity should be 0.3 Osm. Therefore, 0.15%(w/v) glucose is not isotonic with the NaCl solution.
2. 0.55%(w/v) glucose solution:
Following the same calculation as above, the concentration of glucose in moles per liter (M) is:
(0.55 g / 180.16 g/mol) x (1000 mL / 100 mL) = 0.306 M
This solution is approximately isotonic with the NaCl solution as the osmolarity is close to 0.3 Osm.
3. 1.5%(w/v) glucose solution:
Again, using the same calculation as above, the concentration in moles per liter (M) is:
(1.5 g / 180.16 g/mol) x (1000 mL / 100 mL) = 0.083 M
This solution is not isotonic with the NaCl solution because its osmolarity is significantly lower.
4. 5.4%(w/v) glucose solution:
Once again, using the same calculation as above, the concentration in moles per liter (M) is:
(5.4 g / 180.16 g/mol) x (1000 mL / 100 mL) = 0.300 M
This solution is also approximately isotonic with the NaCl solution as the osmolarity is close to 0.3 Osm.
Therefore, the solutions of glucose that are isotonic with a 0.15 M NaCl solution are the 0.55%(w/v) and 5.4%(w/v) glucose solutions.
To determine which of the given glucose solutions is isotonic with a 0.15 M NaCl solution, we need to understand the concept of isotonicity.
Isotonic solutions have the same osmotic pressure as the reference solution. In this case, the reference solution is the 0.15 M NaCl solution. Osmotic pressure is determined by the concentration and the type of solute particles present in a solution.
First, let's find the osmotic pressure of the 0.15 M NaCl solution. The osmotic pressure of a solution can be calculated using the formula:
Osmotic pressure (π) = nRT
Where:
- n is the number of moles of solute particles
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin (usually room temperature, around 25 °C)
Since NaCl dissociates into two particles (Na+ and Cl-) when dissolved in water, the 0.15 M NaCl solution will have an osmotic pressure of:
π = 2 * 0.15 mol/L * 0.0821 L·atm/(mol·K) * 298 K = 7.27 atm
Next, we need to calculate the osmotic pressure of the given glucose solutions using the same formula. However, since glucose does not dissociate like NaCl, each molecule of glucose contributes only one particle.
Now, let's calculate the osmotic pressure for each glucose solution:
1. 0.15%(w/v) glucose solution:
The concentration of glucose is 0.15 g/100 mL. To convert it to Molarity (mol/L), we need to know the molecular weight of glucose, which is 180.2 g/mol. So, the concentration of glucose in Molarity will be:
(0.15 g / 100 mL) / 180.2 g/mol = 8.32 * 10^(-4) M
Therefore, the osmotic pressure for this solution is:
π = 1 * 8.32 * 10^(-4) mol/L * 0.0821 L·atm/(mol·K) * 298 K = 0.019 atm
2. 0.55%(w/v) glucose solution:
The concentration of glucose is 0.55 g/100 mL. Converting it to Molarity:
(0.55 g / 100 mL) / 180.2 g/mol = 3.05 * 10^(-3) M
Calculating the osmotic pressure:
π = 1 * 3.05 * 10^(-3) mol/L * 0.0821 L·atm/(mol·K) * 298 K = 0.069 atm
3. 1.5%(w/v) glucose solution:
The concentration of glucose is 1.5 g/100 mL. Converting it to Molarity:
(1.5 g / 100 mL) / 180.2 g/mol = 8.32 * 10^(-3) M
Calculating the osmotic pressure:
π = 1 * 8.32 * 10^(-3) mol/L * 0.0821 L·atm/(mol·K) * 298 K = 0.190 atm
4. 5.4%(w/v) glucose solution:
The concentration of glucose is 5.4 g/100 mL. Converting it to Molarity:
(5.4 g / 100 mL) / 180.2 g/mol = 2.99 * 10^(-2) M
Calculating the osmotic pressure:
π = 1 * 2.99 * 10^(-2) mol/L * 0.0821 L·atm/(mol·K) * 298 K = 0.838 atm
Based on the calculated osmotic pressures, we can see that the solution with an osmotic pressure closest to the 0.15 M NaCl solution (7.27 atm) is the 0.15%(w/v) glucose solution with an osmotic pressure of 0.019 atm.
Therefore, the correct answer is 0.15%(w/v) glucose solution.