a) [tex]c= \frac{md}{ms} *100=[tex] \frac{3.4}{3.4+36.6}*100= \frac{340}{40}=8.5 % [/tex]
ms=3.4+36.6=40 g sol AgNO3
b)AgNO3 + NaCl -> NaNO3+AgCl
170 g AgNO3.........58.5 g NaCl
3.4..........................x g NaCl
[tex]x= \frac{3.4*58.5}{170}= 1.17 g NaCl[/tex]
[tex] ms_{NaCl} = \frac{1.17*100}{20}= 5.85 g sol NaCl[/tex]
nr moli NaCl = 5.85/58.5=0.1 moli
c)precipitat: AgCl
170 g AgNO3............143.5 g AgCl
3.4 g AgNO3..............y g AgCl
[tex]y= \frac{3.4*143.5}{170} = 2.87 g AgCl [/tex] precipitat