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Which of the following solutions are isotonic (R = 0.0821 atmK-1mol-1)
- a)0.01 M glucose
- b)0.01 M NaNO3
- c)500 mL solution containing 0.3g Urea
- d)0.04 N HCl
Correct answer is option 'A,C'. Can you explain this answer?
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Most Upvoted Answer
Which of the following solutions are isotonic (R = 0.0821 atmK-1mol-1)...
Both ac are correct. As explanation given in the uploaded image by me. Molecular weight of urea is 60. So 500ml 0.3 gm urea solution is 0.01M solution. In this way both are isotonic
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Which of the following solutions are isotonic (R = 0.0821 atmK-1mol-1)...
Isotonic solutions are those solutions that have the same osmotic pressure as the reference solution. The osmotic pressure of a solution depends on the number of solute particles present in it.
To determine which of the given solutions are isotonic, we need to calculate the osmotic pressure of each solution and compare it to the reference solution. The osmotic pressure can be calculated using the formula:
π = nRT/V
Where:
π = osmotic pressure
n = number of moles of solute
R = gas constant (0.0821 atmK-1mol-1)
T = temperature in Kelvin
V = volume of the solution in liters
Let's calculate the osmotic pressure for each solution.
a) 0.01 M glucose:
In this case, the solute is glucose (C6H12O6). To calculate the number of moles of glucose, we can use the formula:
moles = (molarity) x (volume in liters)
Given:
Molarity = 0.01 M
Volume = 1 L (assuming 1 L solution)
moles of glucose = (0.01 M) x (1 L)
moles of glucose = 0.01 mol
Now, let's calculate the osmotic pressure using the formula:
π = (0.01 mol) x (0.0821 atmK-1mol-1) x (298 K) / (1 L)
π ≈ 2.43 atm
b) 0.01 M NaNO3:
In this case, the solute is NaNO3. Following the same procedure as above, we can calculate the osmotic pressure. However, since NaNO3 dissociates into two ions (Na+ and NO3-), we need to consider the total number of particles.
moles of NaNO3 = (0.01 M) x (1 L)
moles of NaNO3 = 0.01 mol
Since NaNO3 dissociates into two particles, the total number of particles is 0.01 mol x 2 = 0.02 mol.
π = (0.02 mol) x (0.0821 atmK-1mol-1) x (298 K) / (1 L)
π ≈ 4.88 atm
c) 500 mL solution containing 0.3 g Urea:
In this case, the solute is urea (CO(NH2)2). Following the same procedure as above, we can calculate the osmotic pressure. However, we need to convert the mass of urea to moles first.
Given:
Mass of urea = 0.3 g
Molar mass of urea = 60.06 g/mol
moles of urea = (0.3 g) / (60.06 g/mol)
moles of urea ≈ 0.005 mol
π = (0.005 mol) x (0.0821 atmK-1mol-1) x (298 K) / (0.5 L)
π ≈ 0.2 atm
d) 0.04 N HCl:
In this case, the solute is HCl. Following the same procedure as above, we can calculate the osmotic pressure. However, since HCl completely dissociates into two ions
To determine which of the given solutions are isotonic, we need to calculate the osmotic pressure of each solution and compare it to the reference solution. The osmotic pressure can be calculated using the formula:
π = nRT/V
Where:
π = osmotic pressure
n = number of moles of solute
R = gas constant (0.0821 atmK-1mol-1)
T = temperature in Kelvin
V = volume of the solution in liters
Let's calculate the osmotic pressure for each solution.
a) 0.01 M glucose:
In this case, the solute is glucose (C6H12O6). To calculate the number of moles of glucose, we can use the formula:
moles = (molarity) x (volume in liters)
Given:
Molarity = 0.01 M
Volume = 1 L (assuming 1 L solution)
moles of glucose = (0.01 M) x (1 L)
moles of glucose = 0.01 mol
Now, let's calculate the osmotic pressure using the formula:
π = (0.01 mol) x (0.0821 atmK-1mol-1) x (298 K) / (1 L)
π ≈ 2.43 atm
b) 0.01 M NaNO3:
In this case, the solute is NaNO3. Following the same procedure as above, we can calculate the osmotic pressure. However, since NaNO3 dissociates into two ions (Na+ and NO3-), we need to consider the total number of particles.
moles of NaNO3 = (0.01 M) x (1 L)
moles of NaNO3 = 0.01 mol
Since NaNO3 dissociates into two particles, the total number of particles is 0.01 mol x 2 = 0.02 mol.
π = (0.02 mol) x (0.0821 atmK-1mol-1) x (298 K) / (1 L)
π ≈ 4.88 atm
c) 500 mL solution containing 0.3 g Urea:
In this case, the solute is urea (CO(NH2)2). Following the same procedure as above, we can calculate the osmotic pressure. However, we need to convert the mass of urea to moles first.
Given:
Mass of urea = 0.3 g
Molar mass of urea = 60.06 g/mol
moles of urea = (0.3 g) / (60.06 g/mol)
moles of urea ≈ 0.005 mol
π = (0.005 mol) x (0.0821 atmK-1mol-1) x (298 K) / (0.5 L)
π ≈ 0.2 atm
d) 0.04 N HCl:
In this case, the solute is HCl. Following the same procedure as above, we can calculate the osmotic pressure. However, since HCl completely dissociates into two ions
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