Correct Answer - a
`16.9 g AgNO_(3)` is present in `100 mL` solution.
`:. 8.45 g AgNO_(3)` is present in `50 mL` solution.
` 5.8 g NaOl` is present in `100 mL` solution.
`:. 2.9 g NaCl` is present in `50 mL` solution.
`AgNO_(3)+NaCl rarr AgCl+ NaNO_(3)`
`8.45/170 mol" " 2.9/58.5`
`=0.049 mol=0.049 mol rarr 0" " 0`
`After "0 " 0 rarr 0.049 mol 0.049`
reaction" " mol
Mass of `AgCl` precipitated
`=0.049xx143.5 g =7 g AgCl`.