The freezing point depression of a 0.01 m aq | vedclass

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

Question

(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 0.0054$,$K_f = 1.80$,and $m = 0.01$. Substi

Vedclass · Accepted Answer

$[Pt(NH_3)_4Cl_2]Cl_2$

Vedclass · Answer

(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 0.0054$,$K_f = 1.80$,and $m = 0.01$. Substituting the values: $0.0054 = i 	imes 1.80 	imes 0.01$. $i = \frac{0.0054}{0.018} = 3$. Since the van't Hoff factor $i = 3$,the complex dissociates into $3$ ions in the solution. Among the options,$[Pt(NH_3)_4Cl_2]Cl_2$ dissociates as $[Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$,which gives $1 + 2 = 3$ ions. Therefore,the correct formula is $[Pt(NH_3)_4Cl_2]Cl_2$.

The freezing point depression of a $0.01 \ m$ aqueous solution of $[Pt(NH_3)_4Cl_4]$ is $0.0054^\circ C$. If $K_f$ for water is $1.80 \ K \ kg \ mol^{-1}$,what is the correct formula for the complex?

Similar Questions

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

$pH$ of a $0.1 \ M$ monobasic acid is found to be $2$. Hence its osmotic pressure at a given temperature $T \ K$ is (in $RT$)

For a $0.1 \, molal$ aqueous solution of $2-$iodopropanoic acid,the degree of dissociation is $5 \%$. The freezing point of the solution will be ............ $^o C$.

Which of the following aqueous solutions has the highest boiling point?

$0.004 \ M$ $Na_2SO_4$ solution is isotonic with $0.01 \ M$ glucose solution,then the degree of dissociation of $Na_2SO_4$ is ........... $\%$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module