Solveeit Logo
Login

Question

Question: What is the vapour pressure at \({100^ \circ }{\text{C}}\) of a solution containing \(18{\text{ g}}\...

What is the vapour pressure at 100∘C{100^ \circ }{\text{C}} of a solution containing 18 g18{\text{ g}} of water and 12.96 g12.96{\text{ g}} sucrose?

Explanation

Solution

To solve this we must know the Raoult’s law. The Raoult’s law states that the partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent measured in the solution by the mole fraction. Calculate the molar mass of sucrose and water. The molecular formula for sucrose is C12H22O11{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}.

Complete solution:
We know Raoult's law. The Raoult’s law states that the partial vapour pressure of a solvent in a solution is equal to the vapour pressure of the pure solvent measured in the solution by the mole fraction.
The expression for Raoult’s law is as follows:
p0−pp0=w×Mm×W\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{w \times M}}{{m \times W}}
Where p0{p^0} is the vapour pressure of the pure solvent,
pp is the vapour pressure of the solution,
ww is the weight of the non-volatile solute,
mm is the molar mass of the non-volatile solute,
MM is the molar mass of the pure solvent,
WW is the weight of the pure solvent.
The molecular formula of sucrose is C12H22O11{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}. Thus, the molar mass of sucrose i.e. non-volatile solute is 342 g/mol342{\text{ g/mol}}. The molar mass of water i.e. pure solvent is 18 g/mol18{\text{ g/mol}}.
The weight of sucrose i.e. non-volatile solute is 12.96 g12.96{\text{ g}}. The weight of water i.e. pure solvent is 18 g18{\text{ g}}.
The vapour pressure of water i.e. pure solvent at 100∘C{100^ \circ }{\text{C}} is 760 mm760{\text{ mm}}.
Thus,
p0−pp0=w×Mm×W\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{w \times M}}{{m \times W}}
p=p0−w×Mm×W×p0p = {p^0} - \dfrac{{w \times M}}{{m \times W}} \times {p^0}
p=760 mm−12.96 g×18 g/mol342 g/mol×18 g×760 mmp = 760{\text{ mm}} - \dfrac{{12.96{\text{ g}} \times 18{\text{ g/mol}}}}{{342{\text{ g/mol}} \times 18{\text{ g}}}} \times 760{\text{ mm}}
p=731.2 mmp = 731.2{\text{ mm}}

Thus, the vapour pressure at 100∘C{100^ \circ }{\text{C}} of a solution containing 18 g18{\text{ g}} of water and 12.96 g12.96{\text{ g}} sucrose is 731.2 mm731.2{\text{ mm}}.

Note: From this we can say that the vapour pressure of a pure solvent decreases when a non-volatile solute is added to it. This is known as lowering in vapour pressure. The vapour pressure lowering is a colligative property of a solution. Lowered vapour pressure results in elevated boiling point.