The minimum volume of water required to dissolve 2. 78 g lead (II) chloride to get a saturated solution
`( K _(sp ) of PbCl _(2) = 3.2 xx 10 ^(-8) ` mlar mass of `PbCl _(2) = 278 g mol ^(-1) )` is -------- L.

Answer

5

Explanation

`underset((S))(Pb Cl _(2) )hArr underset((2s))(Pb ^(2+) )+ 2Cl ^(-)`
`K _(sp ) = [ Pb ^(2+) ] [ Cl ] ^(2) = (S) ( 2 S) ^(2) = 4 S ^(3)`
`therefore 4 S ^(3) = 3.2 xx 10 ^(-8)`
`S = root (3 ) ((3.2 xx 10 ^(-1))/( 4)) = 2 xx 10 ^(-3) mol L ^(-1)`
Now, Molarity = `("No. of moles")/("Volume of solution(L)")`
`therefore` Morality `= ("Mass of substance ")/("Molar mass") `
`xx (1)/("Volume of solution (L)") `
`therefore 2 xx 10 ^(-3) = ( 2 . 78)/( 278) xx (1)/("Volume of solution (L)")`
`therefore` Volume of solution `= (2. 78)/(( 278) xx 2 xx 10 ^(3)) = 5 L `
` therefore` Minimum volume of water required to dissolve `2. 78 gPbCl _(2)` to get saturated solution is 5 L .