Question

Mole fraction iodine in benzene is 0.2. What is the molarity of iodine in benzene

Answer 1

mole fraction = 0.2.
therefore in 100 moles, 20 moles are of solute and 80 moles are of benzene(solvent)
mass of benzene = 80 * Mol.Wt of Benzene = 80* 78 = 6240g
for 6240 g of solvent we have 20 moles of solute
Therefore for 1000g of solvent we have (20)*(1000)/(6240) = 3.20 molal

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Answer 2

Let,

the number of moles of Iodine(solute) in benzene(solvent) be 'x' mol.

and,

The mass of solvent(benzene) be '1000g' or '1kg'

Given:

Mole fraction of Iodine = 0.2

To Find:

Molality of iodine in benzene

Solution:

Molar mass of Benzene=78g

Thus, the number of moles of benzene in 1kg of itself is :

$n = \frac{1000}{78} = 12.821mol$

A.T.Q:

$0.2 = \frac{(x)mol}{(x + 12.821)mol}$

$0.2(x) + (0.2)(12.821) = x$

$1x - 0.2x= 2.564$

$x = \frac{2.564}{0.8}$

$x = 3.21 \: mol$

Now, as

$molality(m) = \frac{no. \: of \: moles \: of \: solute}{mass \: of \: solvent(kg)}$

$(m) \: of \: iodine = \frac{(3.21)mol}{(1)kg}$

.

Answer:

$m = 3.21 \: m \: \: or \: 3.21 \: mol \: kg { }^{ - 1}$