If the relative decrease in vapour pressure i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$100$

Vedclass · Answer

(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 .......... $\%$.

Similar Questions

The freezing point of one molal $NaCl$ solution,assuming $NaCl$ to be $100\%$ dissociated in water,is ............ $^oC$. (molal depression constant $= 1.86$)

Which of the following will have the highest boiling point at $1 \ atm$ pressure?

Osmotic pressure of a $4.44\%$ solution of anhydrous $CaCl_2$ was found to be $16.42 \ atm$ at $27 \, ^\circ C$. What is the degree of dissociation of $CaCl_2$ in $\%$?

The osmotic pressure values of $0.1 \ M$ solutions of $KNO_3$ and $CH_3COOH$ are $P_1$ and $P_2$ respectively. Establish the relationship between them.

For a $0.2 \ m$ aqueous solution of a weak acid $HX$,the degree of ionization is $0.3$. If $K_f$ for water is $1.85 \ K \ kg \ mol^{-1}$,the freezing point of the solution will be close to .......... $^oC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module