(B) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$. Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^

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

(B) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$. Given: $\Delta T_f = 0.0744 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$. Substituting the values: $0.0744 = i \times 1.86 \times 0.01$. $i = \frac{0.0744}{0.0186} = 4$. For complete dissociation of $K_x[Fe(CN)_6]$,the van't Hoff factor $i$ is equal to the number of ions produced,which is $x + 1$. $x + 1 = 4 \Rightarrow x = 3$. Therefore,the molecular formula is $K_3[Fe(CN)_6]$.

Class 12 Chemistry Solutions Colligative properties of electrolyte

Easy

The depression in freezing point for $0.01 \ m$ aqueous solution of $K_{x}[Fe(CN)_6]$ is $0.0744 \ K$. The molal depression constant for solvent is $1.86 \ K \ kg \ mol^{-1}$. If the solute undergoes complete dissociation,what is the correct molecular formula for the solute?

GUJCET 2017 All GUJCET

A
$K_{2}[Fe(CN)_6]$
B
$K_{3}[Fe(CN)_6]$
C
$K[Fe(CN)_6]$
D
$K_{4}[Fe(CN)_6]$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Colligative properties of electrolyte

Previous Year

GUJCET 2017 Paper All GUJCET years →

$17.4\% \ (W/V)$ solution 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 $........... \ \%$.

Medium

View Solution

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

Medium

View Solution

An aqueous solution of a salt $MX_2$ at a certain temperature has a van't Hoff factor of $2$. The degree of dissociation for this solution of the salt is

MediumJEE MAIN 2016

View Solution

At higher altitude,the boiling point of water is $95\,^oC$. The amount of $NaCl$ added to $1\,kg$ of water $(K_b = 0.52\,K\,kg\,mol^{-1})$ in order to raise the boiling point of the solution to $100\,^oC$ (assume $90\%$ ionisation of $NaCl$) is .......... $g$.

Medium

View Solution

In which of the following solutions are ions present?

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The depression in freezing point for $0.01 \ m$ aqueous solution of $K_{x}[Fe(CN)_6]$ is $0.0744 \ K$. The molal depression constant for solvent is $1.86 \ K \ kg \ mol^{-1}$. If the solute undergoes complete dissociation,what is the correct molecular formula for the solute?

Similar Questions

$17.4\% \ (W/V)$ solution 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 $........... \ \%$.

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

An aqueous solution of a salt $MX_2$ at a certain temperature has a van't Hoff factor of $2$. The degree of dissociation for this solution of the salt is

At higher altitude,the boiling point of water is $95\,^oC$. The amount of $NaCl$ added to $1\,kg$ of water $(K_b = 0.52\,K\,kg\,mol^{-1})$ in order to raise the boiling point of the solution to $100\,^oC$ (assume $90\%$ ionisation of $NaCl$) is .......... $g$.

In which of the following solutions are ions present?

Vedclass Test Series

Exam Paper Generator

Online Exam Module