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In a sample of sodium carbonate some sod...

In a sample of sodium carbonate some sodium sulphate is also mixed. 1.25 g of this sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralises 20 mL of `(N)/(10)` sulphuric acid. Calculate the percentage of sodium carbonate in the sample.

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To solve the problem step by step, we will follow the outlined procedure based on the information given in the question. ### Step 1: Determine the amount of sulfuric acid used in equivalents The normality of the sulfuric acid is given as \( \frac{N}{10} \). - Normality (N) = \( \frac{1}{10} \) N - Volume of sulfuric acid used = 20 mL = 0.020 L ...
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In a sample of sodium carbonate, some sodium sulphate is mixed. 2.50 g of this sample is dissolved and the volume made up to 500 mL . 25 mL of this solution neutralises 20 mL of N//10 in the sample.

A sample of sodium carbonate contains sodium also. 1.5 g of the sample is dissolved in water and volume raised to 250 mL. 25 mL of this solution requires 20 mL of (N)/(10)H_(2)SO_(4) solution for neutralisation. Calculate the percentage of sodium carbonate in the sample.

Knowledge Check

  • 5 g sample contain only Na_(2)CO_(3) and Na_(2)SO_(4) . This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H_(2)SO_(4) . Calcalute the % of Na_(2)SO_(4) in the sample .

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    B
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    C
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    D
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    A
    42.4
    B
    57.6
    C
    36.2
    D
    0.576

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