A physiological saline solution is 0.92% NaCl by mass. What is the osmotic pressure of such a solution at a body temperature of 37 $\circ$ C?
A)8.0 atm
B)3.9 atm
C)2.3 atm
D)4.3 atm
E)4.1 atm

Question

Quizplus · Accepted Answer

The osmotic pressure of a solution is given by the van't Hoff equation:$$\pi = MRT$$where:* $\pi$ is the osmotic pressure in atmospheres (atm)* $M$ is the molarity of the solution in moles per liter (mol/L)* $R$ is the ideal gas constant, which is 0.0821 L⋅atm/(mol⋅K)* $T$ is the temperature in Kelvin (K)To solve this problem, we need to first convert the mass percentage of NaCl to molarity. We can do this using the following formula:$$Molarity = \frac{Mass \space percentage \times Density}{Molecular \space weight}$$where:* Mass percentage is the mass of NaCl in grams per 100 grams of solution* Density is the density of the solution in grams per milliliter (g/mL)* Molecular weight is the molecular weight of NaCl in grams per mole (g/mol)The density of a physiological saline solution is approximately 1.00 g/mL, and the molecular weight of NaCl is 58.44 g/mol. Therefore, the molarity of a 0.92% NaCl solution is:$$Molarity = \frac{0.92 \space g/100 \space g \times 1.00 \space g/mL}{58.44 \space g/mol} = 0.0157 \space mol/L$$Now we can plug this value into the van't Hoff equation to calculate the osmotic pressure:$$\pi = MRT = 0.0157 \space mol/L \times 0.0821 \space L⋅atm/(mol⋅K) \times (37 + 273) \space K = 4.1 \space atm$$Therefore, the osmotic pressure of a 0.92% NaCl solution at a body temperature of 37°C is 4.1 atm.

A Physiological Saline Solution Is 0 ∘\circ∘ C? A)8

A Physiological Saline Solution Is 0 $\circ$ C?
A)8