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At 300 K, the osmotic pressure of a decinormal solution of sodium chloride is 4.82 atm.
The degree of dissociation of sodium chloride is x x 10^-2. The value of x is $(R = 0.082 L atm K^{- 1} {mol}^{- 1})$

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answer is B.

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Detailed Solution

$π = iCRT 4.82 = i \times \frac{1}{10} \times 0.082 \times 300 i = 1.96 NaCl ⇌ {Na}^{+} {Cl}^{-} 1 - α α α i = 1 - α + α + α i = 1 + α α = 0.96 degree of dissociation = 0.96 = \underset{x}{\underset{⏟}{96}} \times 10^{- 2}$

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