1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate. Calculate the percentage purity of the sample.

1.5 g of an impure sample of sodium sulphate dissolved in water was tr

1.	1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO_(4) were obtained as dry precipitate. Calculate the percentage purity of the sample.
Answer» Solution :Given : 1.5 g of impure `Na_(2)SO_(4) overset("TREATED with")UNDERSET(BaCl_(2)"gave")rarr 1.7 g` of `BaSO_(4)` The chemical equation representing the REACTION is : `underset(=142 g)underset(2xx23+32+4xx16)(Na_(2)SO_(4))+BaCl_(2)rarr underset(=233 g)underset(137+32+4xx16)(BaSO_(4))` Step 1. To calculate the mass of `Na_(2)SO_(4)` which PRODUCES 1.74 g of `BaSO_(4)`. From the chemical equation, 233 g of `BaSO_(4)` are produced from `Na_(2)SO_(4)=142 g` `therefore` 1.74 g of `BaSO_(4)` are produced from `Na_(2)SO_(4)=(142)/(233)xx1.74=1.06 g` This is the mass of pure `Na_(2)SO_(4)` present in 1.5 g of impure sample. Step 2. To calculate the percentage purity of impure sample. 1.5 g of impure sample contains pure `Na_(2)SO_(4)=1.06 g` `therefore 100 g` of the impure sample will CONTAIN pure `Na_(2)SO_(4)=(1.06)/(1.5)xx100=70.67 g` Thus, percentage purity of impure sample = 70.67.