If #"6.5 g"# of solute #X# is dissolved in #"50 g"# of ether solvent (#"74 g/mol"#) whose vapor pressure dropped from #"445 mm Hg"# to #"410 mm Hg"#, what is the molar mass of #X#?
1 Answer
I got
This is asking you to use Raoult's law to look at vapor pressure reduction from
#P_A = chi_(A(l))P_A^"*"# where:
#P_A# is the vapor pressure of solvent#A# .#"*"# indicates pure solvent.#chi_(A(l)) = (n_A)/(n_A + n_X)# is the mol fraction of#A# in the solution phase.#n_A# is mols of solvent#A# .
Ether is also called diethyl ether, or
Those mols can be found from the mol fraction.
#chi_(A(l)) = P_A/P_A^"*"#
#= ("410 mm Hg")/("445 mm Hg")#
#= n_A/(n_A + n_X) = 0.9213#
We know the mols of the solvent to be:
#n_A = m_A/M_A = 50 cancel"g" xx "1 mol ether"/(74.122 cancel"g")# ,(
#m# being the mass in#"g"# and#M# being the molar mass in#"g/mol"# .)
#=> n_A = # #"0.6746 mols ether"#
So, we can solve for the mols of the solute
#0.9213 = ("0.6746 mols ether")/("0.6746 mols ether" + n_X)#
#=> 0.9213 (0.6746 + n_X) = 0.6746#
#=> 0.6215 + 0.9213n_X = 0.6746#
#=> n_X = "0.05758 mols"#
This means the molar mass of
#color(blue)(M_X) = m_X/n_X = "6.5 g X"/"0.05758 mols X" = color(blue)("112.88 g/mol")#
though you only have two sig figs, apparently.