Name: Expressing Concentration by Mass Percent
Uploaded: 2019-07-19
Duration: 3 min 32 s
Description: Mass percentage is a way of representing the concentration of an element in a compound or a component in a mixture. Let us learn more through this video.

Question

Complete the following table.

Solute	Dissociation/association reaction	Degree of dissociation/association	n	i
$KCl$		$1$
$H_{2} {SO}_{4}$		$1$
${CH}_{3} COOH$ (in water)		$0.2$
${CH}_{3} COOH$ (in benzene)		$0.5$
Urea
$NaBr$		$0.8$
A	$3 A \to A_{3}$	$1$

Accepted Answer

The relation between the degree of dissociation or association of the solute and the Van't Hoff factor $(i)$ is,

For dissociation:

$α = \frac{i - 1}{n - 1}$ .

For association:

$α = \frac{i - 1}{\frac{1}{n} - 1}$ .

Here, ${CH}_{3} COOH$ is the number of ions of the electrolytes and $α$ is the degree of dissociation or association.

The electrolytes $KCl, H_{2} {SO}_{4}, {CH}_{3} {COOH}_{aqueous}$ and $NaBr$ show dissociation and their number of ions is equal to $n$ .

${CH}_{3} COOH$ in benzene exists as a dimer due to hydrogen bonding.

Urea $({NH}_{2} {CONH}_{2})$ is a non-electrolyte. So, $n = 1$ and $i = 1$ .

$\begin{matrix} KCl & H_{2} {SO}_{4} & {CH}_{3} COOH in water & {CH}_{3} COOH in benzene & NaBr & 3 A \to A_{3} \\ α = \frac{i - 1}{n - 1} & α = \frac{i - 1}{n - 1} & α = \frac{i - 1}{n - 1} & α = \frac{i - 1}{\frac{1}{n} - 1} & α = \frac{i - 1}{n - 1} & α = \frac{i - 1}{\frac{1}{n} - 1} \\ 1 = \frac{i - 1}{2 - 1} & 1 = \frac{i - 1}{3 - 1} & 0.2 = \frac{i - 1}{2 - 1} & 0.5 = \frac{i - 1}{\frac{1}{2} - 1} & 0.8 = \frac{i - 1}{2 - 1} & 1 = \frac{i - 1}{\frac{1}{3} - 1} \\ i = 2 & i = 3 & i = 1.2 & i = 0.75 & i = 1.8 & i = 0.33 \end{matrix}$

Solute Dissociation/association reaction Degree of dissociation/association n i

$KCl$ $KCl \to K^{+} + {Cl}^{-}$ $1$ $2$ $2$

$H_{2} {SO}_{4}$ $H_{2} {SO}_{4} \to 2 H^{+} + {SO}_{4}^{2 -}$ $1$ $3$ $3$

${CH}_{3} COOH$ (in water) ${CH}_{3} COOH \to H^{+} + {CH}_{3} COO$ $0.2$ $2$ $1.2$

${CH}_{3} COOH$ (in benzene) $2 {CH}_{3} COOH \to {({CH}_{3} COOH)}_{2}$ $0.5$ $0.5$ $0.75$

Urea No association or dissociation No association or dissociation Not defined $1$

$NaBr$ $NaBr \to {Na}^{+} + Br$ $0.8$ $2$ $1.8$

A $3 A \to A_{3}$ $1$ $0.33$ $0.33$

Complete the following table.

Solute Dissociation/association reaction Degree of dissociation/association n i

$KCl$ $1$

$H_{2} {SO}_{4}$ $1$

${CH}_{3} COOH$ (in water) $0.2$

${CH}_{3} COOH$ (in benzene) $0.5$

Urea

$NaBr$ $0.8$

A $3 A \to A_{3}$ $1$

Important Questions on Solutions

Learn and Prepare for any exam you want !

Complete the following table. Solute Dissociation/association reaction Degree of dissociation/association n i KCl 1 H2SO4 1 CH3COOH (in water) 0.2 CH3COOH (in benzene) 0.5 Urea NaBr 0.8 A 3A→A3 1

Important Questions on Solutions

Learn and Prepare for any exam you want !

Complete the following table.

Solute Dissociation/association reaction Degree of dissociation/association n i

$KCl$ $1$

$H_{2} {SO}_{4}$ $1$

${CH}_{3} COOH$ (in water) $0.2$

${CH}_{3} COOH$ (in benzene) $0.5$

Urea

$NaBr$ $0.8$

A $3 A \to A_{3}$ $1$