What weight of `AgCl` will be precipitated when a solution containing `4.77 g NaCl` is added to a solution of `5.77 g` of `AgNO_(3)`.

Options

A. `4.88 g`
B. `5.77 g`
C. `4.77 g`
D. None of these

Answer

A

Explanation

`{:(,AgNO_(3)+,NaClrarr,AgCl+,NaNO_(3)),(,(5.77)/(170),(4.77)/(58.5),0,0),("Initial mole",0.034,0.082,0,0),("Mole after mixing",0,0.048,0.034,0.034):}`
`therefore` Weight of `AgCl=0.034xx143.5=4.88g`