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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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  1. 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

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