Suppose the osmotic pressure of blood at 37 °C is 7.65 atm and
you are preparing an intravenous (IV) NaCl solution for a
patient. What concentration (in M) of NaCl solution should you
prepare so the resulting IV solution is isotonic with blood, if
the van’t Hoff factor of NaCl is i=1.86?
I have no clue with that one... Please help!!
To determine the concentration of the NaCl solution that would result in an isotonic IV solution, we need to know the definition of an isotonic solution. An isotonic solution is one that has the same osmotic pressure as the fluid it is mixed with. In this case, we want the osmotic pressure of the NaCl solution to match the osmotic pressure of blood.
The formula to calculate the osmotic pressure of a solution is:
Π = i * n * R * T
Where:
Π is the osmotic pressure
i is the van't Hoff factor
n is the concentration of solute in mol/L
R is the ideal gas constant (0.0821 L·atm/(K·mol))
T is the temperature in Kelvin
Let's rearrange the formula to solve for n (concentration):
n = Π / (i * R * T)
Now we can substitute the given values and solve for n:
Π (osmotic pressure of blood) = 7.65 atm
i (van't Hoff factor) = 1.86
R (ideal gas constant) = 0.0821 L·atm/(K·mol)
T (temperature) = 37 °C = 37 + 273.15 = 310.15 K
n = 7.65 atm / (1.86 * 0.0821 L·atm/(K·mol) * 310.15 K)
n ≈ 0.171 mol/L
Therefore, you should prepare a NaCl solution with a concentration of approximately 0.171 M to make it isotonic with blood.
To prepare an isotonic IV solution of NaCl, we need to match the osmotic pressure of blood, which is given as 7.65 atm. The van't Hoff factor (i) of NaCl is given as 1.86.
The formula to calculate the osmotic pressure (π) of a solution is:
π = i * M * R * T
Where:
- π = osmotic pressure (in atm)
- i = van't Hoff factor
- M = molarity (in mol/L)
- R = ideal gas constant (0.0821 L.atm/(mol.K))
- T = temperature (in Kelvin)
First, let's convert the given temperature of 37 °C to Kelvin:
T = 37 °C + 273.15 = 310.15 K
Now, we can rearrange the formula to solve for M (molarity):
M = π / (i * R * T)
M = 7.65 atm / (1.86 * 0.0821 L.atm/(mol.K) * 310.15 K)
Calculating this value, we find:
M ≈ 0.156 M
Therefore, you should prepare a NaCl solution with a concentration of approximately 0.156 M to create an isotonic IV solution that matches the osmotic pressure of blood.