Question

0.002 molar solution of NaCl having degree of dissociation of 90% at 27 degree Celsius has osmotic pressure =

Answer 1

Heyaa..☺️

Here is ur answer..

0.094 bar

Answer 2

Hope this helps

M=0.002

T=27°C = 300K

Degree of dep = 90% = 90/100 = 0.9

For NaCl, i = dod + 1

=0.9 +1

= 1.9

Osmotic pressure= icst

=1.9 × 0.002× 0.0821× 300

=0.0935 atm

=0.0935 × 1.013 bar

= 0.094 bar