how many moles of AgCl2 in 0.717g ?
0.717/143.32 = 0.005 moles
Ag(NO3)2 + 2NaCl = AgCl2 + 2NaNO3
So you had 2*0.005 = 0.01 moles of NaCl
0.01mol/0.01L = 1mole/L = 1M NaCl
An excess of silver nitrate solution was added to 10.0cm3of sodium chloride solution, and 0.717g of silver chloride was precipitated. Calculate the concentration of the sodium chloride solution in mol dm-3.
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