`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to
a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar

Explanation

To solve the problem, we need to calculate the osmotic pressure of a 0.002 molar solution of NaCl with a degree of dissociation of 90% at 27°C. We will use the formula for osmotic pressure, which is given by:

\[ \pi = iCRT \]

where:
- \(\pi\) = osmotic pressure
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