Question

The concentration of NaCl in red blood cells is approximately 598.50 ppm. For a NaCl solution, assume the solution completely dissociates giving a van ’t Hoff factor of 2. Calculate the osmotic pressure of this solution at body temperature (37°C).

Answer 1

We know that

1 ppm = 1 mg/l

= 1 milligram/litre

Given that concentration of NaCl solution is 595.5 ppm

ie 595.50 ppm = 595.5 mg/l

Weight of NaCl = 595.5 mg = 0.5955 g

Molar mass of NaCl = 58.5 g/mol

Molarity of NaCl solution = 0.5955 g *1000/58.5 * 1000 mL

= 0.0102 M

We know that π = iCST

Where i is the Van't Hoff factor = 2

S is the universal solution constant = 0.0821 atm -L/K-mol

T is the absolute temperature = 37°C = 37+273 = 310 K

π = 2*0.0102 mol/L * 0.0821 L-atm/K-mol* 310 K

π = 0.518 atm =5.18*10^-1 atm