Question

How many gram of `Al_(2)(SO_(4))_(3)` are present in `100mL` of `0.15 m` solution of `Al_(2)(SO_(4))_(3)`? The density of solution is `1.4g//mL`.

Answer 1

`100 mL` solution `= 100xx1.4g= 140g` solution
Moles of `Al_(2)(SO_(4))_(3)` in `100 g` solvent
`=(0.15xx100)/(1000)= 0.015`
`:.` Moles of `Al_(2)(SO_(4))_(3)` in `140 g` solvent
`=(140xx0.015)/(100)= 0.021`
`:.` Weight of `Al_(2)(SO_(4))_(3)` in `140 g` solvent
`= 0.021xx342`
`= 7.182 g`
`:.` Weight of solution `= 140+7.182= 147.182 g`
`:. 147.182g` solution `= 7.182 g Al_(2)(SO_(4))_(3)`
`:. 140 g` solution `=(7.182xx140)/(147.182)`
`:. 100 mL` of solution has `6.83 g Al_(2) (SO_(4))_(3)`