MX_{2} dissociates into M^{2+} and X^{-} ions | vedclass

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

Question

(B) The dissociation reaction is: $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$.Initially,let the concentration be $C$. At equilibrium,the concentration

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The dissociation reaction is: $MX_{2} \rightleftharpoons M^{2+} + 2X^{-}$.Initially,let the concentration be $C$. At equilibrium,the concentrations are: $[MX_{2}] = C(1-\alpha)$,$[M^{2+}] = C\alpha$,and $[X^{-}] = 2C\alpha$.The total number of particles (van't Hoff factor $i$) is given by: $i = 1 - \alpha + n\alpha$,where $n$ is the number of ions produced per formula unit.For $MX_{2}$,$n = 3$ ($1$ $M^{2+}$ and $2$ $X^{-}$ ions).Given $\alpha = 0.5$,we have: $i = 1 + (3-1)\alpha = 1 + 2(0.5) = 1 + 1 = 2$.The ratio of the observed depression of freezing point to the theoretical depression (in the absence of dissociation) is equal to the van't Hoff factor $i$.Therefore,the ratio is $2$.

$MX_{2}$ dissociates into $M^{2+}$ and $X^{-}$ ions in an aqueous solution,with a degree of dissociation $(\alpha)$ of $0.5$. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

Similar Questions

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,the change in freezing point of water $(\Delta T_f),$ when $0.01 \ mol$ of sodium sulphate is dissolved in $1 \ kg$ of water,is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $K$).

$12.25 \ g$ of $CH_3CH_2CHClCOOH$ is added to $250 \ g$ of water to make a solution. If the dissociation constant of the above acid is $1.44 \times 10^{-3}$,the depression in the freezing point of water in $^{\circ}C$ is ($K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$)

$0.5 \ m$ aqueous solution of a weak acid $(HX)$ is $20 \%$ ionized. If $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$,the lowering in freezing point of the solution is: (in $K$)

The freezing point depression constant for water is $1.86 \, ^\circ C \, m^{-1}$. If $5.00 \, g$ of $Na_2SO_4$ is dissolved in $45.0 \, g$ of $H_2O$,the freezing point is changed by $3.82 \, ^\circ C$. Calculate the van't Hoff factor $(i)$ for $Na_2SO_4$.

Of the following $0.10 \ m$ aqueous solutions,which one will exhibit the largest freezing point depression?

Vedclass Test Series

Exam Paper Generator

Online Exam Module