
The vapor pressure of water is $23.8mm{\text{ of }}Hg$ at ${25^ \circ }C$. What is the vapor pressure of 2.50 molal ${C_6}{H_{12}}{O_6}$ ?
Answer
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Hint: According to Raoult’s law the vapour pressure of any solution is the product of the Mole fraction of the solvent and the vapor pressure of the pure solution. It can be given as:
${p_{solution}} = {\chi _{solvent}} \times p_{solvent}^0$ -- (1)
Where $p_{solvent}^0$ is the vapor pressure of pure solvent, in this case water.
Complete answer: We are given the concentration of the solution as 2.50 molal solution. Molality can be given as the moles of solute per kilogram of the solvent.
Hence $2.5molal = 2.5moles/kg{\text{ solvent}}$. Which means that 2.5 moles of glucose is mixed in 1 kilogram of solvent.
Number of moles of solute = 2.5 moles
Mass of the solvent = 1 kg = 1000 g
Number of moles of Solvent $ = \dfrac{{mass}}{{Molar{\text{ Mass}}}} = \dfrac{{1000}}{{18}}moles$
The mole fraction of solvent water can be given as:
${\chi _{water}} = \dfrac{{{n_{{H_2}O}}}}{{{n_{{H_2}O}} + {n_{glucose}}}} = \dfrac{{\dfrac{{1000}}{{18}}}}{{\dfrac{{1000}}{{18}} + 2.5}}$
${\chi _{water}} = \dfrac{{1000}}{{1045}}$
Now, to find the vapor pressure of solution, we need to have the vapor pressure of pure water which is given as $23.8mm{\text{ of }}Hg$
Substituting the values in equation (1)
${p_{solution}} = \dfrac{{1000}}{{1045}} \times 23.8$
${p_{solution}} = 22.775mm{\text{ of }}Hg$
Therefore, the vapor pressure of 2.50 molal ${C_6}{H_{12}}{O_6}$ solution is 22.775 mm of Hg.
Note:
The vapor pressure reduces with the addition of a solute, in this case glucose. This happens because the solvent molecules are displaced by the molecules of the solute. The solvent molecules on the surface are also displaced by the solute particles. Since the no. of solvent molecules on the surface reduces, the no. of molecules that evaporate from the surface also reduces. This lowers the vapor pressure. The addition of solute also increases the boiling point, as more temperature is required now to make the atmospheric pressure to be equal to the vapor pressure of the solution. Relative Lowering of vapor pressure is a colligative property.
${p_{solution}} = {\chi _{solvent}} \times p_{solvent}^0$ -- (1)
Where $p_{solvent}^0$ is the vapor pressure of pure solvent, in this case water.
Complete answer: We are given the concentration of the solution as 2.50 molal solution. Molality can be given as the moles of solute per kilogram of the solvent.
Hence $2.5molal = 2.5moles/kg{\text{ solvent}}$. Which means that 2.5 moles of glucose is mixed in 1 kilogram of solvent.
Number of moles of solute = 2.5 moles
Mass of the solvent = 1 kg = 1000 g
Number of moles of Solvent $ = \dfrac{{mass}}{{Molar{\text{ Mass}}}} = \dfrac{{1000}}{{18}}moles$
The mole fraction of solvent water can be given as:
${\chi _{water}} = \dfrac{{{n_{{H_2}O}}}}{{{n_{{H_2}O}} + {n_{glucose}}}} = \dfrac{{\dfrac{{1000}}{{18}}}}{{\dfrac{{1000}}{{18}} + 2.5}}$
${\chi _{water}} = \dfrac{{1000}}{{1045}}$
Now, to find the vapor pressure of solution, we need to have the vapor pressure of pure water which is given as $23.8mm{\text{ of }}Hg$
Substituting the values in equation (1)
${p_{solution}} = \dfrac{{1000}}{{1045}} \times 23.8$
${p_{solution}} = 22.775mm{\text{ of }}Hg$
Therefore, the vapor pressure of 2.50 molal ${C_6}{H_{12}}{O_6}$ solution is 22.775 mm of Hg.
Note:
The vapor pressure reduces with the addition of a solute, in this case glucose. This happens because the solvent molecules are displaced by the molecules of the solute. The solvent molecules on the surface are also displaced by the solute particles. Since the no. of solvent molecules on the surface reduces, the no. of molecules that evaporate from the surface also reduces. This lowers the vapor pressure. The addition of solute also increases the boiling point, as more temperature is required now to make the atmospheric pressure to be equal to the vapor pressure of the solution. Relative Lowering of vapor pressure is a colligative property.
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