In an aqueous solution of NaCl,the mole fract | vedclass

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(C) The relative lowering in vapour pressure is given by the formula: $\frac{P^o - P_s}{P^o} = i \cdot x_{solute}$.Given that the mole fraction of the

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$0.2$

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(C) The relative lowering in vapour pressure is given by the formula: $\frac{P^o - P_s}{P^o} = i \cdot x_{solute}$.Given that the mole fraction of the solvent $(x_{solvent})$ is $0.9$,the mole fraction of the solute $(x_{solute})$ is $1 - 0.9 = 0.1$.Since $NaCl$ is a strong electrolyte,it dissociates as $NaCl ightarrow Na^+ + Cl^-$,so the van't Hoff factor $(i)$ is $2$.Substituting the values: $\frac{P^o - P_s}{P^o} = 2 	imes 0.1 = 0.2$.

In an aqueous solution of $NaCl$,the mole fraction of the solvent is $0.9$. Then,the relative lowering in vapour pressure will be ......... $atm$.

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