Name: Importance of Chemistry in Everyday Life
Uploaded: 2019-07-19
Duration: 3 min 4 s
Description: Chemicals play a very important role in our daily life and several biological processes. They also find application in the agricultural, textile and medicina...

Question

0.5 g

of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution.

0.5 g

of

{BaSO}_{4}

was obtained as precipitate. Calculate the percentage purity of sample.

Accepted Answer

$0.5 g$ of an impure sample of sodium sulphate is treated with access amount of barium chloride and $0.5 g$ of barium sulphate obtained as white precipitate form.

The chemical reaction is given by,
${Na}_{2} {SO}_{4} + {BaCl}_{2} \to 2 NaCl + {BaSO}_{4} ↓$ .

Here, it is clearly shown that $1$ mole of sodium sulphate produces $1$ mole of barium sulphate. The molecular weight of ${Na}_{2} {SO}_{4} = (23 \times 2) + 32 + (4 \times 16) = 46 + 32 + 64 = 142 g {mol}^{- 1}$ .

The molecular weight of ${BaSO}_{4} = 137 + 32 + 64 = 233 g {mol}^{- 1}$ . So, $233 g$ of ${BaSO}_{4}$ is produced by $142 g$ of ${BaSO}_{4}$ . Therefore, $0.5 g {NaSO}_{4}$ produces $= (\frac{142}{233}) \times 0.5 = 0.305 g$ .

But, the given mass of ${Na}_{2} {SO}_{4} = 0.5 g$ . So, impurity in sample $= (0.5 - 0.305) g = 0.195 g$ . So, purity of sample $= \frac{0.305}{0.500} \times 100 = 61 %$ .

$0.5 g$ of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution. $0.5 g$ of ${BaSO}_{4}$ was obtained as precipitate. Calculate the percentage purity of sample.

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0.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution. 0.5 g of BaSO4 was obtained as precipitate. Calculate the percentage purity of sample.