Question

What is the vapor pressure of the pure solvent if the vapor pressure of a solution of 10 g of sucrose (C6H12O6) in 100 g of ethanol (C2H6O) is 55 mmHg?

Answer 1

Step-by-step explanation:

The given data is as follows.

Vapor pressure of the solution
`(P_(solution))` = 55 mm Hg

Mass of sucrose = 10 g

Molar mass of sucrose = 180 g/mol

Therefore, moles of sucrose present into the solution will be calculated as follows.

No. of moles =
`(mass)/(molar mass)`

=
`(10 g)/(180 g/mol)`

= 0.055 mol

Mass of ethanol is given as 100 g and its molar mass is 46 g/mol.

Hence, number of moles of ethanol will be calculated as follows.

No. of moles =
`(mass)/(molar mass)`

=
`(100 g)/(46 g)`

= 2.174 mol

As mole fraction =
`(no. of moles)/(total number of moles)`

Hence, mole fraction of etahnol will be calculated as follows.

`X_(ethanol)` =
`(no. of moles)/(total number of moles)`

=
`(2.174)/(2.174 + 0.055)`

= 0.975

Now, using Raoult's Law as follows.

`P_(solution) = X_(ethanol) * P_(ethanol)`

`P_(ethanol)` =
`(P_(solution))/(X_(ethanol))`

=
`\frac{55 mm Hg}}{0.975}}`

= 56.4 mm Hg

Thus, we can conclude that the vapor pressure of the pure solvent is 56.4 mm Hg.

Answer 2

56.4 mmHg

Step-by-step explanation:

Given:

Vapor pressure of the solution, P solution = 55 mmHg

The mass of sucrose (C₆H₁₂O₆) = 10 g

Also, Molar mass of sucrose (C₆H₁₂O₆) = 180 g/mol

So, moles = Given mass/ molar mass

Hence, moles of sucrose in the solution = 10 g / 180 g/mol = 0.05556 mol

Given that: Mass of ethanol = 100 g

Molar mass of ethanol = 46 g/mol

Hence, moles of ethanol = 100 g / 46 g/mol = 2.174 mol

Mole fraction of solvent, ethanol is:

X ethanol = 2.174 mol / (2.174 + 0.05556) mol = 0.975

Applying Raoult's Law

P solution = X ethanol*P° ethanol

=> P° ethanol = P solution / X ethanol = 55 mmHg / 0.975 = 56.4 mm Hg