What is the molarity of `SO_(4)^(2-)` ions in aqueous solution that contains 34.2 ppm of `Al_(2)(SO_4)_(3)`? (Assume complete dissociation and density of solution `1 g"/"mL)`

Options

A. `3xx 10 ^(-4) M`
B. `2 xx 10 ^(-4) M`
C. `10 ^(-4) M`
D. `2 xx 10 ^(-5)M`

Answer

A

Explanation

`("mas of " Al _(2) (SO _(4)) _(3))/( "mass of water") xx 10 ^(6) = 34.2`
1L solution contains 1000 g of water `implies` In 1 L solution, mass of `Al _(2) (SO _(4) ) _(3) = ( 34.2 xx 1000)/(10 ^(6)) = 34.2 ms`
Molarity of `Al _(2) ( SO _(4)) _(3) = ( 34. 2 xx 10 ^(-3))/( 342) M = 1 0 ^(-4)`
`Al _(2) underset(10 ^(-4) M )((SO_(4)) _(3))(aq) to underset(2 x 10^(-4) M )(2 Al ^(3+) (aq) + ) underset(3 xx 10 ^(-4) M) (3SO_(4) ^(2-)(aq)), [SO_(4) ^(2-)] = 3 xx 10 ^(-4) M`