Question

The vapour pressure of pure water is 12.3 k Pa at 300 k. Find the vapour pressure of a solution containing 1 molal non volatile salute.​

Answer 1

Answer:

Vapour pressure of the solution = 12.0825 kPa

Explanation:

Given:

• Vapour pressure of pure water = 12.3 kPa
• The solution contains 1 molal non volatile solute

To Find:

• Vapour pressure of the solution.

Solution:

We know that relative lowering of vapour pressure is given by,

$\tt \dfrac{(p_1)^0-p_1}{(p_1)^0} =\dfrac{n_2}{n_1+n_2}$

where p₁⁰ is the vapour pressure of the pure solvent, p₁ is the vapour pressure of the solution, n₁ = number of moles of solvent and n₂ = number of moles of solute.

Number of moles of water = Given mass/Molar mass

Given that the solution contains 1 molal solute, ie, 1000g of the solvent contains 1 mole of solute.

Hence,

Number of moles of water = 1000/18 = 55.56 moles

Number of moles of solute = 1 mole

Substitute the data in the above equation,

$\tt \dfrac{12.3-p_1}{12.3} =\dfrac{1}{1+55.56}$

$\tt \dfrac{12.3-p_1}{12.3} =\dfrac{1}{56.56}$

$\tt 695.688-56.56\: p_1=12.3$

$\tt 56.56\: p_1=683.388$

$\tt p_1=\dfrac{683.388}{56.56}$

$\tt p_1=12.0825\: kPa$

Hence the vapour pressure of the solution is 12.0825 kPa.