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

1 Answer
Sam
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\ %#

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