Question

The van't hoff factor for the dilute aqueous solutiin of hcn (10% dissociated) is

Answer 1

Given:

HCN is 10% dissociated in an dilute aqueous solution.

To find:

Van't Hoff Factor for HCN.

Calculation:

Van't Hoff Factor can be calculated by following these steps:

$HCN \: \longrightarrow \: {H}^{ + } + \: {CN}^{ - } \\ \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ (1 - x) \: \: \: \: \: \: \: \: \: \: x \: \: \: \: \: \: \: \: \: \: \: \: \: x$

Putting x = 10% = 0.1

$(1 - 0.1) \: \: \: \: \: \: 0.1 \: \: \: \: \: \: \: \: \: 0.1$

Let Van't Hoff Factor be denoted by i.

$\therefore \: i = \dfrac{ (1 - x) + x + x}{1}$

$= > \: i = \dfrac{ 1+ x}{1}$

$= > \: i = \dfrac{ 1+ 0.1}{1}$

$= > \: i = 1.1$

So , final answer is:

Van't Hoff Factor for HCN (10% dissociation) is 1.1.

Answer 2

Given:

HCN is 10% dissociated in an dilute aqueous solution.

To Find:

We have to find the van't hoff for the dilute aqueous solution.

Solution:

By considering the given data,

$HCN --- > H^{+}+CN^{-}$

1                       0         0

(1-x)                   x         x

On putting x=10%=0.1 we get,

(1-0.1)                0.1       0.1

Calculating the van't hoff $i$,

$i=\frac{1-x+x+x}{1}\\ i=\frac{1+x}{1}\\ i=\frac{1+0.1}{1}\\ i=1.1$

Hence the value of van't hoff factor is $1.1$

#SPJ2