✒️VAPOR PRESSURE LOWERING
[tex]\tt{\huge{\blue{Solution:}}}[/tex]
The problem wants us to calculate the vapor pressure of a solution. First, we will identify the quantities used in the problem. The quantities used are:
(Note: Sucrose is the solute and water is the solvent.)
[tex]n_{solt} =[/tex] number of moles of solute
[tex]n_{solv} =[/tex] number of moles of solvent
[tex]m_{solt} =[/tex] mass of solute
[tex]m_{solv} =[/tex] mass of solvent
[tex]\text{MM}_{solt} =[/tex] molar mass of solute
[tex]\text{MM}_{solv} =[/tex] molar mass of solvent
[tex]X_{solv} =[/tex] mole fraction of solvent
[tex]P_{solv} =[/tex] vapor pressure of pure solvent
[tex]P_{soln} =[/tex] vapor pressure of solution
Second, we will identify the given values in the problem. These are
[tex]m_{solt} =[/tex] 200 g
[tex]m_{solv} =[/tex] 350 g
[tex]\text{MM}_{solt} =[/tex] 342 g/mol
[tex]\text{MM}_{solv} =[/tex] 18 g/mol
[tex]P_{solv} =[/tex] 17.5 mmHg
Third, we need to find the number of moles of solute and solvent. These are
[tex]n_{solt} = \dfrac{m_{solt}}{\text{MM}_{solt}}[/tex]
[tex]n_{solt} = \dfrac{\text{200 g}}{\text{342 g/mol}}[/tex]
[tex]n_{solt} =[/tex] 0.585 mol
and
[tex]n_{solv} = \dfrac{m_{solv}}{\text{MW}_{solv}}[/tex]
[tex]n_{solv} = \dfrac{\text{350 g}}{\text{18 g/mol}}[/tex]
[tex]n_{solv} =[/tex] 19.444 mol
Fourth, we will find the mole fraction of solvent. The mole fraction of solvent is
[tex]X_{solv} = \dfrac{n_{solv}}{n_{solt} \: + \: n_{solv}}[/tex]
[tex]X_{solv} = \dfrac{\text{19.444 mol}}{\text{0.585 mol + 19.444 mol}}[/tex]
[tex]X_{solv} =[/tex] 0.971
Finally, we can now calculate the vapor pressure of the solution. Therefore, the vapor pressure of the solution is
[tex]P_{soln} = X_{solv}P_{solv}[/tex]
[tex]P_{soln} = (0.971)(\text{17.5 mmHg})[/tex]
[tex]\boxed{P_{soln} = \text{17.0 mmHg}}[/tex]
[tex]\\[/tex]
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