Correct Answer - A
Molarity (M) of `AgNO_(3)=(16.9xx10)/(169.3)=1.0M`
Molarity (M) of `NaCl=(5.8)/(58.5)xx10=0.9914M`
`AgNO_(3) +NaCl to AgCl darr+NaNO_(3)`
Moles `n={:((0.99Mxx50)/(1000),(0.99xx50)/(1000) ),(cong0.05,cong0.05):}`
`therefore` mols of AgCl formed=0.05
Mass of AgCl=0.05`xx142.5=7g`