Question: The vapour pressure in mm Hg of an aqueous solution obtained by adding 18g of glucose to 180g of wat...

Question

The vapour pressure in mm Hg of an aqueous solution obtained by adding 18g of glucose to 180g of water at $100{}^\circ C$ is

Answer 1

Solution

When glucose is added to pure water then a lowering of vapour pressure is observed. The relative lowering of the vapour pressure is equal to the mole fraction of the solute.
We will use the given formula:
$\dfrac{P{}^\circ -P}{P{}^\circ }\,=\,mole\,fraction\,of\,glu\cos e$
Where, $P{}^\circ$ is the vapour pressure of pure water at $100{}^\circ C$

Complete step by step solution: Let’s look at the solution to the given question.
It is given in the question,
Mass of glucose added = 18g
Mass of water = 180g
Molar mass of glucose = 180g
Molar mass of water = 18g
Now, we will calculate the mole fraction of glucose:
Moles of a substance $=\dfrac{given\,mass}{molar\,mass}$
Moles of glucose $=\dfrac{18}{180}\,=\,0.1$
Moles of water $=\dfrac{180}{18\,}\,=\,10$
Now, moles of glucose $=\dfrac{moles\,of\,glu\cos e}{total\,moles\,of\,solution}$
Therefore. Moles of glucose $=\dfrac{0.1}{10+0.1}=\,0.0099$
Now, we will calculate the vapour pressure of solution using the relation
$\dfrac{P{}^\circ -P}{P{}^\circ }\,=\,mole\,fraction\,of\,glu\cos e$ ………..equation 1
Where, $P{}^\circ$ is the vapour pressure of pure water at $100{}^\circ C$
$P{}^\circ$ = 760 mm Hg
On putting the values in equation 1, we get:
$\dfrac{760-P}{760}\,=\,0.0099$
Therefore, P = 752.4 mm Hg
Hence, the answer of the given question is option (D)

Note: The vapour pressure of a substance is defined as the pressure exerted by the vapours of the substance. The relative lowering of vapour pressure is governed by Raoult’s law. According to Raoult’s law the relative lowering of the vapour pressure is equal to the mole fraction of the solute added to the solution.