If the relative decrease in vapour pressure i | vedclass

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(C) The relative lowering of vapour pressure is given by Raoult's law for non-volatile solutes: $\frac{P^o - P_S}{P^o} = \frac{i 	imes n}{i 	imes n

Vedclass · Accepted Answer

$100$

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(C) The relative lowering of vapour pressure is given by Raoult's law for non-volatile solutes: $\frac{P^o - P_S}{P^o} = \frac{i 	imes n}{i 	imes n + N}$.Given: $\frac{P^o - P_S}{P^o} = 0.4$,$n (NaCl) = 1 \, mol$,$N (H_2O) = 3 \, mol$.Substituting the values: $0.4 = \frac{i 	imes 1}{i 	imes 1 + 3}$.Solving for $i$: $0.4(i + 3) = i \implies 0.4i + 1.2 = i \implies 0.6i = 1.2 \implies i = 2$.For $NaCl ightleftharpoons Na^+ + Cl^-$,the van't Hoff factor $i = 1 + \alpha$.$2 = 1 + \alpha \implies \alpha = 1$,which is $100 \%$.

If the relative decrease in vapour pressure is $0.4$ for a solution containing $1 \, mol$ of $NaCl$ in $3 \, mol$ of $H_2O$,the degree of ionization of $NaCl$ is .......... $\%$.

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