Osmotic pressure of 0.1 M aqueous solution o | vedclass

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(A) The formula for osmotic pressure is $\pi = i 	imes C 	imes R 	imes T$.Given: $\pi = 4.92 \ atm$,$C = 0.1 \ M$,$R = 0.0821 \ L \ atm \ K^{-1} \

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(A) The formula for osmotic pressure is $\pi = i 	imes C 	imes R 	imes T$.Given: $\pi = 4.92 \ atm$,$C = 0.1 \ M$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$.Substituting the values: $4.92 = i 	imes 0.1 	imes 0.0821 	imes 300$.$4.92 = i 	imes 2.463$.$i = \frac{4.92}{2.463} \approx 1.998 \approx 2$.For $MgCl_2$,the dissociation is $MgCl_2 ightarrow Mg^{2+} + 2Cl^-$,so $n = 3$.The degree of ionization $\alpha$ is given by $\alpha = \frac{i-1}{n-1}$.$\alpha = \frac{1.998-1}{3-1} = \frac{0.998}{2} = 0.499$.Percentage ionization $= \alpha 	imes 100 = 49.9 \% \approx 50 \%$. Given the options,$49 \%$ is the closest value.

Osmotic pressure of $0.1 \ M$ aqueous solution of $MgCl_2$ at $300 \ K$ is $4.92 \ atm$. What will be the percentage ionization of the salt? $.......... \ \%$

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