If the mole fraction of NaCl in an aqueous solution is 0.0927, what is the percent by mass of NaCl?

1 Answer
May 24, 2016

%\ by \ mass= 24.9\ %

Explanation:

X_(NaCl)=0.0927

X_(NaCl) + X_(H_2O) = 1

X_(H_2O) = 1 - X_(NaCl)

X_(H_2O) = 1 - 0.0927

X_(H_2O) = 0.9073

X_(NaCl) /X_(H_2O) = 0.0927/0.9073=0.1022

X_(NaCl) / X_(H_2O)=( n_(NaCl)/(n_(NaCl) +n_(H_2O)))/( n_(H_2O)/(n_(NaCl) +n_(H_2O)))=( n_(NaCl)/cancel((n_(NaCl) +n_(H_2O))))/( n_(H_2O)/cancel((n_(NaCl) +n_(H_2O))))=n_(NaCl)/n_(H_2O)

n_(NaCl)/n_(H_2O)=0.1022

%\ by \ mass= m_(NaCl)/((m_(NaCl)+m_(H_2O)))xx100

mass= nxxMM

"Where "MM" is the molar mass and "n" is the number of moles."

%\ by \ mass=((n_(NaCl)xxMM_(NaCl)))/((n_(NaCl)xxMM_(NaCl)+n_(H_2O)xxMM_(H_2O)))xx100

"Divide both, numerator and denominator by " n_(H_2O)

%\ by \ mass=(((n_(NaCl)xxMM_(NaCl))/n_(H_2O)))/ (((n_(NaCl)xxMM_(NaCl)+n_(H_2O)xxMM_(H_2O))/n_(H_2O)))xx100

%\ by \ mass=((n_(NaCl)/n_(H_2O)xxMM_(NaCl)))/ ((n_(NaCl)/n_(H_2O)xxMM_(NaCl)+n_(H_2O)/n_(H_2O)xxMM_(H_2O)))xx100

%\ by \ mass=((n_(NaCl)/n_(H_2O)xxMM_(NaCl)))/ ((n_(NaCl)/n_(H_2O)xxMM_(NaCl)+cancel(n_(H_2O))/cancel(n_(H_2O))xxMM_(H_2O)))xx100

%\ by \ mass=((n_(NaCl)/n_(H_2O)xxMM_(NaCl)))/ ((n_(NaCl)/n_(H_2O)xxMM_(NaCl)+MM_(H_2O)))xx100

%\ by \ mass=(0.1022xx58.45 \ g*mol^-1)/ (0.1022xx58.45\ g*mol^1+18.016\ g.mol^-1)xx100

%\ by \ mass= 24.9\ %