Question

7.
A solution is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is
0.9.
a) What is the molality (m) of the solution?
b) Water is added to the above solution such that the mole fraction of water in the solution
becomes 0.9 What is the molality (m) of the solution?

Answer 1

Answer:

a) Molality of the solution = 2.415 m

b) Molality = 6.173 m

Explanation:

Given:

• Mole fraction of ethanol present in the mixture = 0.9
• When water is added to the solution, mole fraction of the water becomes 0.9

To Find:

• Molality of the solution in the first case
• Molality of the solution when water is added to it

Solution:

a) Given that mole fraction of ethanol is 0.9

Hence mole fraction of water present in the mixture = 1 - 0.9 = 0.1

Since the mole fraction of ethanol is greater, ethanol is the solvent and water is the solute

Now molality is given by,

$\boxed{\sf Molality=\dfrac{Number\:of\:moles\:of\:solute}{Weight\:of\:solvent\:in\:kg} }$

We know that,

Number of moles = Given mass/Molar mass

Mass of ethanol (CH₃CH₂OH) = 0.9 × (12 + 3 + 12 + 2 + 16 + 1)

⇒ 0.9 × 46

⇒ 41.4 gm = 0.0414 kg

Substitute in the above equation,

$\sf Molality = \dfrac{0.1}{0.0414}$

$\sf \implies 2.415\:m$

b) Now water is added to the solution that it's mole fraction becomes 0.9

Hence mole fraction of ethanol = 1 - 0.9 = 0.1

Here water is the solvent and ethanol is the solute.

Mass of water = 0.9 × 18 = 16.2 gm = 0.0162 kg

Hence molality is given by,

$\sf Molality=\dfrac{0.1}{0.0162}$

$\sf \implies 6.173\:m$

Answer 2

Answer :-

1]

Ethanol = 0.9

Water present in mixture = 0.1

Molality = 0.9 × (12 + 3 + 12 + 2 + 16 + 1)

Molality = 0.9 × 46

Molality = 41.4 gm/1000 = 0.0414 kg

Molality = 0.1/0.414 = 2.415 m

2]

mole fraction of ethanol = 1 - 0.9 = 0.1

Finding mass of water

0.9 × 18 = 16.2 gm = 0.0162 kg

Molality = 0.1/0.0162 kg

= 6.173