Question

An aqueous solution is 0.182 M NaCl and at 25 C has a density of 1.062 g mL -1. The observed osmotic pressure of the solution is 8.456 atm at 25 C. What fraction of NaCl exists as ion pairs in this solution?

Answer 1

we have formula P = i x M x R x T

where P is osmotc pressure = 8.456 atm , T = 25C = 25+273 = 298 K , M = molarity = 0.182M

and R = gas constant = 0.08206 litere atm/molK , i = vantoff factor , we find i

8.456 atm = i x 0.182M x 0.08206 liter atm/molK x 298 K

i = 1.9 = average number of ions produced when NaCl dissociated

when NaCl dissociates completly we get 2 ions ,   s i = 1.9 meas NaCl dissociated = 0.19/2 = 0.95 fraction

The fraction of NaCl undissociated or existing as pairs = 1-0.95 = 0.05