Question

What is the osmolarity of a 0.8% solution of sodium chloride (NaCl)? Atomic weights: (Na = 23, Cl = 35.5)

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.0 g/mL

use:

mass = density * volume

= 1 g/mL *1*10^3 mL

= 1*10^3 g

This is mass of solution

mass of NaCl = 0.8 % of mass of solution

= 0.8*1000.0/100

= 8.0 g

Molar mass of NaCl,

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

= 1*22.99 + 1*35.45

= 58.44 g/mol

mass(NaCl)= 8.0 g

use:

number of mol of NaCl,

n = mass of NaCl/molar mass of NaCl

=(8 g)/(58.44 g/mol)

= 0.1369 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 0.1369/1

= 0.1369 M

This is molarity of NaCl

In NaCl, there are 2 particles (Na+ and Cl-)

So,

osmolarity = 2*[NaCl]

= 2*0.1369 M

= 0.274 M

Answer: 0.274 M