If 0.2 mole NaCl is added to 2 mole water then vapour pressure of the solution is(vapour pressure of pure water is 760 torr)?

1 Answer
Truong-Son N.
Jun 6, 2018

Well, first of all, the vapor pressure of pure water is #"23.76 torr"# at #25^@ "C"# (it's not the atmospheric pressure unless the water is boiling, which is unstated).

The vapor pressure of an ideal solution is given by Raoult's law:

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

where:

  • #chi_(A(l)) = n_A/(n_A + n_B + . . . )# is the mol fraction of solvent #A# in the solution phase.
  • #P_A# is the vapor pressure of the solution, and #"*"# indicates pure solvent.

The vapor pressure is then:

#color(blue)(P_A) = ("2 mol H"_2"O")/("0.2 mol NaCl" + "2 mol H"_2"O") cdot "23.76 torr"#

#=# #color(blue)("21.60 torr")#

Whatever vapor pressure you use for the solvent, it will be weighted by the mol fraction of the solvent, which in this case is #0.909cdots#.

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