a sample of caustic soda contains 74.6 Na2O by weight. find the purity of the sample
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Let amount of Caustic soda (NaOH) is taken 100g
In 100g of Caustic soda contains 74.6 g of Na₂O .
Let see reaction of Caustic soda :
2NaOH ⇔Na₂O + H₂O
Here it is clear that 2 moles of NaOH form 1 mole of Na₂O
∴ 2(23 + 16 + 1) = 80g of NaOH form (2×23+16) = 62g of Na₂O
∴ 1g of Na₂O = 80/62 g of NaOH
∴ 74.6g of Na₂O = 80/62 × 74.6 = 96.26g
Now , purity of Caustic soda = theoretical amount of NaOH/taken amount of NaOH × 100
= 96.26/100 × 100
= 96.26 %
In 100g of Caustic soda contains 74.6 g of Na₂O .
Let see reaction of Caustic soda :
2NaOH ⇔Na₂O + H₂O
Here it is clear that 2 moles of NaOH form 1 mole of Na₂O
∴ 2(23 + 16 + 1) = 80g of NaOH form (2×23+16) = 62g of Na₂O
∴ 1g of Na₂O = 80/62 g of NaOH
∴ 74.6g of Na₂O = 80/62 × 74.6 = 96.26g
Now , purity of Caustic soda = theoretical amount of NaOH/taken amount of NaOH × 100
= 96.26/100 × 100
= 96.26 %
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