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

May 24, 2016

%\ by \ mass= 24.9\ %

Explanation:

${X}_{N a C l} = 0.0927$

${X}_{N a C l} + {X}_{{H}_{2} O} = 1$

${X}_{{H}_{2} O} = 1 - {X}_{N a C l}$

${X}_{{H}_{2} O} = 1 - 0.0927$

${X}_{{H}_{2} O} = 0.9073$

${X}_{N a C l} / {X}_{{H}_{2} O} = \frac{0.0927}{0.9073} = 0.1022$

${X}_{N a C l} / {X}_{{H}_{2} O} = \frac{{n}_{N a C l} / \left({n}_{N a C l} + {n}_{{H}_{2} O}\right)}{{n}_{{H}_{2} O} / \left({n}_{N a C l} + {n}_{{H}_{2} O}\right)} = \frac{{n}_{N a C l} / \cancel{\left({n}_{N a C l} + {n}_{{H}_{2} O}\right)}}{{n}_{{H}_{2} O} / \cancel{\left({n}_{N a C l} + {n}_{{H}_{2} O}\right)}} = {n}_{N a C l} / {n}_{{H}_{2} O}$

${n}_{N a C l} / {n}_{{H}_{2} O} = 0.1022$

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

$m a s s = n \times M M$

$\text{Where "MM" is the molar mass and "n" is the number of moles.}$

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

$\text{Divide both, numerator and denominator by } {n}_{{H}_{2} O}$

 %\ 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\ %