Determine the mass in grams of glucose (180.2 g/mol) needed to prepare a solution with a vapor pressure of 40.8 mmHg. The glucose is dissolved in 575g of water at 35C. The vapor pressure of pure water at 35C is 42.2 mmHg.

Please help me, I don't know what to start with or even what equation this requires

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asked by lex
  1. Psoln = Xsolvent*Posolvent
    Substitute and solve for Xsolute

    Then Xsolute = mols solvent)/(mols solvent)+(mols solute)
    Solve for mols solute and convert to grams solute. Post your work if you get stuck.

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  2. Given VP(H₂O) = 42.2-mmHg @35ᵒC & Needed VP(Soln) = 40.8-mmHg
    VP(Solution) = VP(solvent) – ΔVP(solvent)
    ΔVP(solvent) = X(solute)∙VP(solvent)

    (42.2-mmHg – 40.8-mmHg) = [(m/180.2)/(m/180.2) + (575/18)]∙42.2-mmHg ; m = mass glucose (g)
    Solving for mass of glucose => *m = 198.2-grams needed to effect a drop in VP of 1.4-mmHg in solvent VP.

    *Verify by substituting m = 198.2-g into ΔVP = [(m/180.2)/(m/180.2) + (575/18)]∙42.2-mm => 1.4-mmHg drop in VP of solvent => Final VP(Solution) = 42.2-mmHg – 1.4-mmHg = 40.8-mmHg.

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    posted by Doc48

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