Question

Mole fraction of I2 in CCl4 is 0.2, then molality value? which of the following not correct​

Answer 1

Given:

Mole fraction of I2 in CCl4 is 0.2.

To find:

Molality of I2.

Calculation:

Let the solution be x molal in strength.

So, there will be x moles of Iodine present in 1000 g of solvent (CCl4).

So, moles of CCl4

$\therefore\:\sf{moles \: of \: CCl_{4} = \dfrac{1000}{12 + (35.5 \times 4)} }$

$= > \sf{moles \: of \: CCl_{4} = \dfrac{1000}{154} }$

$= > \sf{moles \: of \: CCl_{4} = 6.49 \: moles}$

Now , we can say that :

$\therefore \: \sf{( \chi)_{I_{2}} = \dfrac{x}{x + 6.49} }$

$= > \: \sf{0.2 = \dfrac{x}{x + 6.49} }$

$= > \: \sf{x + 6.49 = \dfrac{x}{0.2} }$

$= > \: \sf{x + 6.49 = \dfrac{10x}{2} }$

$= > \: \sf{x + 6.49 = 5x}$

$= > \: \sf{4x = 6.49 }$

$= > \: \sf{x = 1.62 \: m }$

So, final answer is:

$\boxed{ \: \bf{molality = 1.62 \: m }}$