Question

Vapour pressure of a water at 293K is $ 17.55 $ mmHg. Calculate the vapour pressure of water at $ 293K $ when $ 25g $ of glucose is dissolved in $ 450g $ of water?

Answer 1

Hint: To solve this question we have to take in consideration the molar masses of the water and the glucose molecule, then taking into account the molar mass we will find the number of males of each entity, after this we will calculate the mole fraction of the glucose molecule and will relate it with a vapour pressure.

Complete Step By Step Answer:
As we studied the approach to solve this question, we will proceed first by mentioning the molar mass of glucose and water and then finding the number of moles of each.
So, let’s begin:
As we know, the molar mass of glucose is $ 180gmo{l^{ - 1}} $ and the molar mass of water is $ 18gmo{l^{ - 1}} $ .
Now finding the number of moles:
So number of moles of glucose is $ = \dfrac{{25}}{{180}} = 0.139 $
Also, number of moles of water is $ = \dfrac{{450}}{{18}} = 25 $
Now, we will find the mole fraction of the glucose molecule:
So, the mole fraction of the glucose will be $ = \dfrac{{0.139}}{{0.139 + 25}} = 0.0055 $
Also we know that the vapour pressure lowering is directly proportional to the mole fraction:
So our relation between vapour pressure and the mole fraction is:
 $ \Rightarrow $ $ = \dfrac{{{P_o} - P}}{{{P_0}}} = X $
 $ \Rightarrow $ $ = \dfrac{{17.535 - P}}{{17.535}} = 0.0055 $
 $ \Rightarrow $ $ P = 17.438 $
Hence, our answer is $ P = 17.438 $
Hence, on the basis of our above discussion and solution: So, the vapour pressure of the solution is $ \Rightarrow $ $ P = 17.438 $
Step by step follow these simple procedures.

Note:
Vapour pressure, pressure exerted by a vapour when the vapour is in equilibrium with the liquid or solid form, or both, of the same substance. This vapour pressure is also governed by Raoult's law.