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

Answer:

56.4 mmHg

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

Answer 2

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 = $\frac{mass}{molar mass}$

                                   = $\frac{10 g}{180 g/mol}$

                                   = 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 = $\frac{mass}{molar mass}$

                                  = $\frac{100 g}{46 g}$

                                  = 2.174 mol

As mole fraction = $\frac{no. of moles}{total number of moles}$

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

            $X_{ethanol}$ = $\frac{no. of moles}{total number of moles}$

                                              = $\frac{2.174}{2.174 + 0.055}$

                                              = 0.975

Now, using Raoult's Law  as follows.

              $P_{solution} = X_{ethanol} \times P_{ethanol}$

             $P_{ethanol}$ = $\frac{P_{solution}}{X_{ethanol}}$

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

                                    = 56.4 mm Hg

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