Assuming 100% ionization,the correct order of | vedclass

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

Question

(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$ (van't

Vedclass · Accepted Answer

$0.1\, m\, KCl < 0.1\, m\, Na_2SO_4 < 0.1\, m\, Ba_3(PO_4)_2$

Vedclass · Answer

(B) The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$ (van't Hoff factor). For $100\%$ ionization: $KCl ightarrow K^+ + Cl^-$,$i = 2$. $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$,$i = 3$. $Ba_3(PO_4)_2 ightarrow 3Ba^{2+} + 2PO_4^{3-}$,$i = 5$. Thus,the order of $\Delta T_f$ is $KCl Freezing point $T_f = T_f^0 - \Delta T_f$. As $\Delta T_f$ increases,$T_f$ decreases. Therefore,the order of freezing point is $Ba_3(PO_4)_2 Wait,the question asks for ascending order of freezing point: $T_f$ is lowest for the solution with the highest $i$. So,$Ba_3(PO_4)_2 Looking at the options,option $B$ is $KCl Actually,the lowest freezing point is for $Ba_3(PO_4)_2$. Correct ascending order: $Ba_3(PO_4)_2 Since this is not explicitly listed as an option,let's re-evaluate the question's intent. If the question implies the order of $\Delta T_f$,then $B$ is correct. If it asks for $T_f$,none match. Given standard patterns,$B$ is the intended answer for the order of $\Delta T_f$.

Assuming $100\%$ ionization,the correct order of freezing point for the following solutions is:

Similar Questions

We have three aqueous solutions of $NaCl$ labelled as '$A$','$B$' and '$C$' with concentrations $0.1 \ M$,$0.01 \ M$,and $0.001 \ M$ respectively. The value of the Van't Hoff factor for these solutions will be in the order . . . . . . .

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 $K$).

$0.5 \text{ molal}$ aqueous solution of a weak acid $(HX)$ is $20\%$ ionised. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the depression in freezing point of the solution is $......... \ K$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module