A 0.002 , M solution of NaCl having a degree | vedclass

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(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{i-1}{n-1}$,where $n$ is the number of ions produced per formula unit of $NaCl$ $(n

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$0.094 \, bar$

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(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{i-1}{n-1}$,where $n$ is the number of ions produced per formula unit of $NaCl$ $(n=2)$.Given $\alpha = 0.9$,we have $0.9 = \frac{i-1}{2-1}$,which gives the van't Hoff factor $i = 1.9$.The osmotic pressure $\pi$ is calculated using the formula $\pi = i 	imes C 	imes R 	imes T$.Here,$C = 0.002 \, M$,$R = 0.082 \, L \cdot bar \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300 \, K$.Substituting the values: $\pi = 1.9 	imes 0.002 	imes 0.082 	imes 300 = 0.09348 \, bar \approx 0.094 \, bar$.

$A$ $0.002 \, M$ solution of $NaCl$ having a degree of dissociation of $90 \%$ at $27 \, ^{\circ}C$ has an osmotic pressure equal to:

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