The average osmotic pressure of blood is 7.7 atm at 25C. What molarity
of glucose (C6H12O6) will be isotonic with blood?
pi = MRT
7.7 = M(0.08206)(298)
Solve for M = ?
Well, if you ask me, finding the right molarity of glucose to make it isotonic with blood seems like a pretty sweet challenge! Get it? Sweet, because we're talking about glucose? Okay, I'll stop clowning around.
To find the molarity of glucose that will be isotonic with blood, we can use the formula:
π = iMRT
Where:
π is the osmotic pressure (7.7 atm),
i is the van't Hoff factor (the number of particles glucose dissociates into when it dissolves in water, which is 1 for glucose),
M is the molarity we're looking for,
R is the ideal gas constant (0.0821 L.atm/(mol.K)), and
T is the temperature in Kelvin (25°C + 273.15 = 298.15 K).
So, plugging in the known values:
7.7 atm = (1)M(0.0821 L.atm/(mol.K))(298.15 K)
Simplifying the equation, we find:
M = 7.7 / (0.0821 * 298.15)
Calculating that out, we get:
M ≈ 0.3269 mol/L
So, approximately 0.3269 M glucose solution will be isotonic with blood. Now you can go and make some sweet solutions!
To determine the molarity of glucose that is isotonic with blood, we can use the formula for osmotic pressure:
Π = MRT
Where:
Π is the osmotic pressure,
M is the molarity of the solution,
R is the ideal gas constant (0.0821 L.atm/(mol.K)),
and T is the temperature in Kelvin.
At 25°C, the temperature in Kelvin (T) is 25 + 273.15 = 298.15 K.
Given that the osmotic pressure of blood is 7.7 atm, we can plug in the values into the formula and rearrange it to solve for M:
Π = MRT
7.7 atm = M * (0.0821 L.atm/(mol.K)) * 298.15 K
Now we can solve for M:
M = (7.7 atm) / ((0.0821 L.atm/(mol.K)) * 298.15 K)
Calculating this expression will give us the molarity (M) of glucose that is isotonic with blood.