A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normality of the solution is 0.1 N. The value of x is ……….. (The atomic mass of Na is 23 g/mol)

Equivalent of solute = 0.1 × 0.1 Mole of solute (Na2CO3.xH2O) = [0.1 × 0.1]1/2 Mass of Na2CO3.xH2O = [0.1 × 0.1] 1/2 x [106 + 18x] = 1.43 => [106 + 18x] = 286 18x =180 x = 10

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A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normality of the solution is 0.1 N.

Chemistry

Last Post by nikhil04 4 years ago

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19/09/2020 10:41 am

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aditi07

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A 100 mL solution was made by adding 1.43 g of Na₂CO₃.xH₂O. The normality of the solution is 0.1 N. The value of x is ………..

(The atomic mass of Na is 23 g/mol)

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19/09/2020 10:53 am

nikhil04

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Equivalent of solute = 0.1 × 0.1

Mole of solute (Na₂CO₃.xH₂O) = [0.1 × 0.1]1/2

Mass of Na₂CO₃.xH₂O = [0.1 × 0.1] 1/2 x [106 + 18x] = 1.43

=> [106 + 18x] = 286

18x =180

x = 10

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A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normality of the solution is 0.1 N.

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