Question

If an aqueous solution is 2.17% (w/v) in sodium chloride, NaCl , what is the osmolarity of the solution?

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

This is mass of solution

mass of NaCl = 2.17 % of volume of solution

= 2.17*1000/100

= 21.7 g

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 21.7 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(21.7 g)/(58.44 g/mol)

= 0.3713 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 0.3713/1

= 0.3713 M

This is concentration of NaCl

NaCl dissociates into 2 particles

So, i = 2

Osmolarity = i * [NaCl]

= 2*0.3713 osM

= 0.743 osM

Answer: 0.743