Raoult’s Law Question 6

The vapour pressure of an aqueous solution of glucose at 373K is 0.96 atm. Calculate the molarity of the solution.

In order to be able to find the vapour pressure of the solution the mole fraction of both the water and the glucose.

Since the concentration needs to be found, it will be assumed that x grams of glucose are dissolved in 1litre of water.

1 mole water = 18g and 1 mole glucose = 180g

Therefore the moles of each component are:

$moles\;water = \frac{1000}{18}=44.44\;moles$

$moles\;glucose = \frac{x}{180}$

This means that the mole fraction of water can be found:

$mf_{(water)}=\frac{moles_{(water)}}{total\;moles}=\frac{44.44}{44.44+\frac{x}{180}}$

This would make the vapour pressure of the solution as: $VP_{(solution)}=mf_{(water)}*VP{(pure\;water)}$

$0.96=\frac{44.44}{44.44+\frac{x}{180}}*1atm$

VP_water is equivalent to 1atm since this is the vapour pressure at boiling point.

Making x subject of the formula becomes 333.3g.

This would make the moles of glucose as:

$Moles\;of\;glucose= \frac{333.3}{180}=1.852\;moles$

And therefore the concentration would be equivalent to 1.852 mole dm^-3