Correct Answer - C
Plan For the calculation of mass of AgCl precipitated , we find mass of `AgNO_(3)` and NaCl in equal volume with the help of mole concept.
`16.9%` solution of `AgNO_(3)` means 16.9 g `AgNO_(3)` is present in 100 mL solution.
`:.8.45 g AgNO_(3)` will be present in 50 mL solution. Similarly,
5.8 g NaCl is present in 50 mL solution
`{:(,AgNO_(3),+,NaCl,to,AgCl,+,NaNO_(3)),("Initial mole",(8.45)/(169.8),,(2.9)/(58.5),," "0,," "0),(,=0.049,,=0.049,,,,),("After reaction"," "0,,0,,0.049,,0.049):}`
`:.` Mass of AgCl precipitated
`=0.049xx143.5=7 g`