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. `3xx10^(-4) M`
B. `2xx10^(-4) M`
C. `10^(-4) M `
D. `2xx10^(-5)M`

Answer

A

Explanation

`("mass of" Al_(2)(SO_4)_(3))/("mass of water")xx10^(6)=34.2`
1L solution contains `1000 g` of water `implies` In 1 L solution, mass of `Al_(2)(SO_4)=(34.2xx1000)/(10^6)=34.2 mg`
Molarity of `Al_(2)(SO_4)_(3)=(34.2xx10^(-3))/(342)M=10^(-4)`
`{:(Al_(2)(SO_4)_(3)(aq) to 2Al^(3+)(aq)+3SO_(4)^(2-)(aq)),(10^(-4)M" "2xx10^(-4)" "3xx10^(-4)M):}," "[SO_(4)^(2-)]= 3xx10^(-4)M`.