Correct Answer - 1

`16.9%` solution of `AgNO_(3)` implies that every 100 mL of solution contain `16.9 gAgNO_(3)`. Thus, `50 mL` of solution will contain `8.45 g AgNO_(3)`.

Similarly, `5.8% NaCl` solution implies that every `100 mL` of solution contains `5.8 g NaCl`. Thus, `50 mL` of solution will contain `2.9 g NaCl`.

`n_(AgNO_(3))=(mass_(AgNO_(3)))/("molar mass"_(AgNO_(3)))`

`=(8.45 g)/(170 g mol^(-1))=0.049 mol`

`n_(NaCl)=(mass_(NaCL))/("molar mass"_(NaCl))`

`=(2.9 g)/(5.85" g "mol^(-1))=0.049 mol`

According to the balanced chemical equation

`AgNO_(3)(aq)+NaCl(aq) rarr AgCl(s)+NaNO_(3)(aq)`

`Mass_(AgCl)" precipitated"=(n//AgCl)("molar mass"_(AgCl))`

`=(0.049 mol)(143.5 g mol^(-1))`

`=7.0 g`