Question

has been solved by taking
34. Vapour pressure of water at 293 K is 17.535 mm Hg.
Calculate the vapour pressure of water at 293 K when
25 g of glucose is dissolved in 450 g of water.
25 o​

Answer 1

Answer:

Vapour pressure of the solution = 17.438 mm Hg

Explanation:

Given:

• Vapour pressure of water at 293 K = 17.525 mm Hg
• Mass of glucose = 25 g
• Mass of water = 450 g

To Find:

• Vapour pressure of the solution

Solution:

Molar mass of the solute (C₆H₁₂O₆) = 180 g/mol

Molar mass of the solvent (water) = 18 g/mol

The relative lowering of vapour pressure is given by,

$\boxed{\sf \dfrac{(p_1)^0-p_1}{(p_1)^0}=\dfrac{w_2\times M_1}{w_1\times M_2} }$

where p₁⁰ is the vapour pressure of the pure solvent, p₁ is the vapour pressure of the solution, w₁ is the weight of solvent, w₂ is the weight of the solute, M₁ is the molar mass of solvent, M₂ is the molar mass of solute

Substituting the data we get,

$\sf \dfrac{17.535-p_1}{17.535} =\dfrac{25\times 18}{450\times 180}$

$\sf \dfrac{17.535-p_1}{17.535} =\dfrac{450}{450\times 180}$

$\sf \dfrac{17.535-p_1}{17.535} =\dfrac{1}{ 180}$

Cross multiplying we get,

$\sf 3156.3-180\:p_1=17.535$

$\sf 180\:p_1=3138.765$

$\sf p_1=\dfrac{3138.765}{180}$

$\sf p_1=17.438\:mm\:Hg$

Hence the vapour pressure of the solution is 17.438 mm Hg.

Answer 2

Solution :-

We know that

$\sf \dfrac{p_1 - p1}{p_1} = \dfrac{Mw_{1}}{Mw_2}$

$\sf \dfrac{17.535 - p1}{17.535}=\dfrac{25\times18}{450 \times 180}$

$\sf \dfrac{17.535 - p1}{17.535}=\dfrac{450}{81000}$

$\sf\dfrac{17.535 - p1}{17.535} = \dfrac{1}{180}$

$\sf 180(17.535 - p1) = 1(17.535)$

$\sf 31563 - 180p1 = 17.535$

$\sf 180p1= 3156.3 + 17.535$

$\sf 180p1=3138.765$

$\sf p1 = \dfrac{3138.765}{17.535}$

$\sf p1=17.44 mm/Hg$