Question bank/JEE Main/Chemistry
JEE MainChemistryModeratePhysical Chemistry
The hydrated salt Na_2CO_3.xH_2O undergoes 63 loss in mass on
heating and becomes anhydrous.Calculate the value of x.
Na_2CO_3.xH_2O 63 x
Show answer and solution
[IMAGE] Let 100 g of Na2CO3. xH2O be present
[IMAGE] mole of H2O formed 6678 + 1134 x = 1800 x
666 x = 6678
[IMAGE] gy- [IMAGE] Na2CO3. xH2O [IMAGE] 100 g
[IMAGE] H2O 6678 + 1134 x = 1800 x
666 x = 6678
[IMAGE] mole of H2O formed 6678 + 1134 x = 1800 x
666 x = 6678
[IMAGE] gy- [IMAGE] Na2CO3. xH2O [IMAGE] 100 g
[IMAGE] H2O 6678 + 1134 x = 1800 x
666 x = 6678
Source: Grass-G / CH-04 Paper 02 Chemistry with Hindi 22-8-2020.docx
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