How many grams of sucrose must be added to 320 g of water to lower the vapor pressure by `1.5 mmHg` at `25^(@)C`? The vapor pressure of water is 23.8 mmHg and the molar mass of sucrose is `342 g mol^(-1)`.
Strategy: If the solute like sucrose is nonvolatile, the vapor pressure of the solution is entirely due to the vapor pressure of the solvent, `P_("solution")=P_("solvent")`. According to Raoult's law, `P_("soln")=P_("solv")^(0) chi_("solv")`, which can be rearranged to the form `chi_("solv")=P_("soln")//P_("solv")^(0)`. This equation can then be solved to find the number of moles of sucrose and hence the grams.

Explanation

Step 1. Calculate the valor pressure of the solution, `P_("soln")`.
Subtract the amount of vapour pressure lowering from the vapour pressure of the pure solvent, `P_("solv")`:
`P_("soln")=P_("solv")^(0)-DeltaP_(1)`
`=23.8 mmHg-1.5 mmHg`
`=22.3 mmHg`
Step 2. Calculate the mole fraction of water, `chi_("solv")`.
Since `P_("soln")=P_("solv")^(0) chi_("solv")`
then
`chi_("solv")=P_("soln")/P_("solv")^(0)=(22.3 mmHg)/(23.8 mmHg)=0.937`
Step 3. Calculate moles of sucrose
Mole fraction of water is the number of moles of water divided by the total number of moles of sucrose plus water:
`chi_("solv")=n_(H_(2)O)/n_("Total")`
Since the number of moles of water is
`n_(H_(2)O)=(mass_(H_(2)O))/(molar mass_(H_(2)O))`
`=(320 g)/(18 g mol^(-1))`
`=17.8 mol`
then the total number of moles of solution is
`n_("total")=n_(H_(2)O)/chi_("solv")=(17.8 mol)/0.937`
`=9.0 mol`
Subtracting the number of moles of `H_(2)O` from the total numbe of moles gives the number of moles of sucrose needed:
`n_("sucrose")=n_("total")-n_(H_(2)O)`
`=19.0 mol- 17.8 mol`
`=1.2 mol`
Step 4. Calculate mass of sucrose
Converting moles into grams then gives the mass of sucrose needed:
`mass_("sucrose")=(n_("sucrose"))(molar mass_("sucrose"))`
`=(1.2 mol)(342.0 g mol^(-1))`
`=410.4 g`
`=4.1xx10^(2) g`
Alternatively, we can solve for `chi_(solv)` by using equation `(2.55)`:
`(DeltaP)/P_(1)^(0)=chi_(2)`
`(1.5 mm Hg)/(23.8 mm Hg)=chi_(2)=0.063`
`chi_(1)=1- chi_(2)=0.937`