Relative decrease in vapour pressure of an aqueous NaCl is 0.167. Number of moles of NaCl present in 180 g of H20 is? 1) 2 Mol 2) 1mol 3) 3 Mol 4) 4 mol

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Feb 11, 2018

Here's a way to do it without knowing the multiple choice answers.

I got #"2.003 mols NaCl"#. You have given two sig figs, so we can say #"2.0 mols NaCl"#.


The decrease in vapor pressure of the solvent #A# is given by:

#DeltaP_A = P_A - P_A^"*"#

Assuming the solution follows Raoult's law (i.e. that it is ideal), we get:

#DeltaP_A = chi_(A(l))P_A^"*" - P_A^"*"#

#= (chi_(A(l)) - 1)P_A^"*"#

#= -chi_(B(l))P_A^"*"#

where #chi_(B(l))# is the mol fraction of the solute in the solution phase, #P_A# is the vapor pressure of the solvent in the context of the solution, and #"*"# indicates pure solvent.

In this form, it is clear that the change is always negative. If the relative decrease is #0.167#, then we say that

#|DeltaP_A|/P_A^"*" = 1 - P_A/P_A^"*" = chi_(B(l)) = 0.167#

and we see how this turns out to be a fraction; as we have just shown, #P_A < P_A^"*"# necessarily, whenever any nonvolatile solute is added to the solvent, so #0 < (|DeltaP_A|)/(P_A^"*") < 1#.

By definition,

#chi_("NaCl"(l)) = (n_("NaCl"(l)))/(n_("H"_2"O"(l)) + n_("NaCl"(l)))#

This means that the mols of #"NaCl"# in the solution phase is found from:

#n_("NaCl"(l)) = 0.167(n_("H"_2"O"(l)) + n_("NaCl"(l)))#

The mols of water are:

#n_("H"_2"O"(l)) = "180 g H"_2"O" xx ("1 mol")/("18.015 g H"_2"O") = "9.992 mols"#

Therefore:

#"9.992 mols" + n_("NaCl"(l)) = n_"tot"#

and:

#n_("NaCl"(l)) = 0.167("9.992 mols" + n_("NaCl"(l)))#

Solving for the mols of #"NaCl"# in the solution phase, we get:

#0.167("9.992 mols") = (1 - 0.167)n_("NaCl"(l))#

#color(blue)(n_("NaCl"(l))) = (0.167/(1-0.167))("9.992 mols")#

#=# #color(blue)("2.003 mols")#

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Feb 11, 2018

Answer:

Use Raoult's Law.
2) 2 moles

Explanation:

Math details here:
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/Solutions_and_Mixtures/Colligative_Properties/Vapor_Pressure_Lowering

#X_"solvent" = n_"water"/(n_"NaCl" + n_"water")#

#DeltaP = (X_"solvent" − 1) xx P_"solvent"^0#

#(180g H_2O)/((18g)/"mol") = 10 mol H_2O#

#X_"solvent" = 10/(n_"NaCl" + 10)#

For 2 Moles #NaCl#:
#X_"solvent" = 10/(2 + 10) = 0.833#

#DeltaP = (X_"solvent" − 1) xx P_"solvent"^0#

Relative change #= 0.167 = 1 - 0.833#

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