Solution.
x = n(NaCl)
y = n(KCl)
M(NaCl) = 58.44 g/mol
M(KCl) = 74.55 g/mol
n(AgNO3)=C×V=0.012mol
58.44×x+74.55×y=0.770
x+y=0.012
x = 0.0077 mol
y = 0.0043 mol
m(NaCl) = 0.45 g
m(KCl) = 0.32 g
w(NaCl)=0.770.45×100%=58.44%
w(KCl)=0.770.32×100%=41.56%
Answer:
w(NaCl) = 58.44 %
w(KCl) = 41.56 %
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