Which of the following 0.10 m aqueous soluti | vedclass

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

Question

(A) The freezing point depression is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since the molality $(m)$ and the solvent $(K_f)$ are t

Vedclass · Accepted Answer

$Al_2(SO_4)_3$

Vedclass · Answer

(A) The freezing point depression is given by the formula $\Delta T_f = i 	imes K_f 	imes m$. Since the molality $(m)$ and the solvent $(K_f)$ are the same for all solutions,the freezing point depends on the van't Hoff factor $(i)$. $Al_2(SO_4)_3$ dissociates as $2Al^{3+} + 3SO_4^{2-}$,giving $i = 5$. $C_5H_{10}O_5$ is a non-electrolyte,$i = 1$. $KI$ dissociates as $K^+ + I^-$,giving $i = 2$. $C_{12}H_{22}O_{11}$ is a non-electrolyte,$i = 1$. Since $Al_2(SO_4)_3$ has the highest van't Hoff factor $(i=5)$,it will show the maximum depression in freezing point,resulting in the lowest freezing point.

Which of the following $0.10 \ m$ aqueous solution will have the lowest freezing point?

Similar Questions

The correct relationship between the boiling points of very dilute solutions of $AlCl_3$ $(t_1)$ and $CaCl_2$ $(t_2)$,having the same molar concentration is

$12.25 \ g$ 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 $1.86 \ K \ kg \ mol^{-1}$)

Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

Which one of the following $0.10 \ M$ aqueous solutions will exhibit the largest freezing point depression?

Observe the following statements:
Statement-$I$: The boiling point of $0.1 \ M$ urea solution is less than that of $0.1 \ M$ $KCl$ solution.
Statement-$II$: Elevation of boiling point is inversely proportional to molar mass of solute.
The correct answer is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module