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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Stoichiometry
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m.viddya (101)

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Q.The hydrated salt Na2SO4 .nH2O undergoes 56% loss in weight on heating and becomes anhydrous.The value of "n" is?
1.5
2.7
3.4
4.10
    
Greatdreams (3307)

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Write the equation and equate the molecular masses you

will surely get the answer to be 10


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BALGANESH (682)

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Molecular weight of Na2SO4 .NH2O =23.2 +32 +16.4+ n(2+16)
                                                              =142 +18n
 
Now since it loses 56% 0n becoming anhydrous ,therefore weight of water in the compound  is 56 %  =  (56 /100) 142 + 18n =  18n
 
On solving n = 10
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m.viddya (101)

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hey!can ya juz explain how u derived dat eq..56/100*142+18n=18?????
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