The number of moles of lead nitrate needed to coagulate 2 mole of colloidal `[AgI]I^(-)` is :

Options

A. 2
B. 1
C. `1//2`
D. `2//3`

Answer

B

Explanation

`2[AgI]I^(-)+Pb^(2+) to PbI_(2) +2AgI`
Thus 2 mole of `[AgI]I^(-)` has two mole -ve charges, it is equivalent to total charge on `Pb^(2+)` of one mole of `Pb(NO_(3))_(2)`.