The percentage dissociation of a salt (MX_3) | vedclass

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

Question

(D) The dissociation reaction for the salt is: $MX_3 \rightarrow M^{3+} + 3X^{-}$.Here,the number of ions produced per formula unit is $n = 1 + 3 = 4$

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(D) The dissociation reaction for the salt is: $MX_3 ightarrow M^{3+} + 3X^{-}$.Here,the number of ions produced per formula unit is $n = 1 + 3 = 4$.The relationship between the van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is given by: $i = 1 + (n - 1)\alpha$.Substituting the given values: $2 = 1 + (4 - 1)\alpha$.$2 = 1 + 3\alpha$.$3\alpha = 1$.$\alpha = \frac{1}{3} \approx 0.3333$.Therefore,the percentage dissociation is $0.3333 	imes 100 = 33.33\%$.The nearest integer is $33\%$.

The percentage dissociation of a salt $(MX_3)$ solution at a given temperature $($van't Hoff factor $i = 2)$ is $..........\%$ ($Nearest$ $integer$).

Similar Questions

$A$ $0.0020 \ m$ aqueous solution of an ionic compound $[Co(NH_3)_5(NO_2)]Cl$ freezes at $-0.00732 \ ^oC.$ Number of moles of ions which $1 \ mol$ of ionic compound produces on being dissolved in water will be
$(K_f = 1.86 \ ^oC/m)$

The Van't Hoff factor for sodium phosphate would be

$17.4\% (W/V)$ of potassium sulphate $(mol. wt. = 174)$ is isotonic with $4\% (W/V)$ aqueous solution of $NaOH$. If $NaOH$ is $100\%$ ionised,the degree of ionisation of potassium sulphate is $.......... \%$.

The freezing points of equimolar solutions of glucose,$KNO_3$ and $AlCl_3$ are in the order of:

The molal elevation constant of water is $0.51 \ K \, kg \, mol^{-1}$. The boiling point of $0.1 \ m$ aqueous $NaCl$ solution is nearly .......... $^oC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module