If 0.1 M solution of NaCl is isotonic with 1 | vedclass

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(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi)$.For $NaCl$ solution: $\pi_1 = i 	imes C_1 	imes R 	imes T$.For urea s

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$0.83$

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(B) Two solutions are isotonic if they have the same osmotic pressure $(\pi)$.For $NaCl$ solution: $\pi_1 = i 	imes C_1 	imes R 	imes T$.For urea solution: $\pi_2 = C_2 	imes R 	imes T$ (since $i = 1$ for urea).Given $1.1 \ w \%$ urea solution means $1.1 \ g$ of urea in $100 \ mL$ of solution, so $C_2 = \frac{1.1 \ g / 60 \ g \ mol^{-1}}{0.1 \ L} = 0.1833 \ M$.Equating the osmotic pressures: $i 	imes 0.1 = 0.1833$.$i = 1.83$.For $NaCl ightarrow Na^+ + Cl^-$, the number of ions $n = 2$.The degree of ionization $\alpha$ is given by $\alpha = \frac{i - 1}{n - 1}$.$\alpha = \frac{1.83 - 1}{2 - 1} = 0.83$.

If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$

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