2.013 g of a commercial sample of NaOH containing sodium carbonate as an impurity was dissolved to give 250 ml solution. A 10 ml portion of this solution required 20 ml of 0.1 N sulphuric acid for complete nuetralisation. Calculate % by weight of sodium carbonate.

Question

Umakant biswal · Answer

@ pranjal
i am giving below the correct approach for this query
20 mL of 0.1N sulphuric acid will require 2 milli equivalents of NaOH for complete neutralisation,
Hence, in 10 mL of the solution, 2 milli equivalents of NaOH are present
So, in 250 mL of solution, there will be 50 milli equivalents of NaOH, which will have a mass of 40g*0.050 = 2g
The remaining 2.013-2 = 0.013g will be of sodium carbonate.
So, the percentage weight of sodium carbonate in the sample will be = (0.013/2.013)*100 = 0.6458%
HOPE IT CLEARS YOUR DOUBT
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2.013 g of a commercial sample of NaOH containing sodium carbonate as an impurity was dissolved to give 250 ml solution. A 10 ml portion of this solution required 20 ml of 0.1 N sulphuric acid for complete nuetralisation. Calculate % by weight of sodium carbonate.

2.013 g of a commercial sample of NaOH containing sodium carbonate as an impurity was dissolved to give 250 ml solution. A 10 ml portion of this solution required 20 ml of 0.1 N sulphuric acid for complete nuetralisation. Calculate % by weight of sodium carbonate.

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2.013 g of a commercial sample of NaOH containing sodium carbonate as an impurity was dissolved to give 250 ml solution. A 10 ml portion of this solution required 20 ml of 0.1 N sulphuric acid for complete nuetralisation. Calculate % by weight of sodium carbonate.

2.013 g of a commercial sample of NaOH containing sodium carbonate as an impurity was dissolved to give 250 ml solution. A 10 ml portion of this solution required 20 ml of 0.1 N sulphuric acid for complete nuetralisation. Calculate % by weight of sodium carbonate.

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