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NA 2 SO 4 XH 2 O ON HEATING SUFFERS A LOSS OF WEIGHT OF 56% THEN FIND X , MOL.WT ?
Na2SO4.XH2O -> Na2SO4 + X H2O (BECOMES ANHYDROUS) 1MOLE == 1 == X MOLES (== MEANS REACT) THUS, 100/(142+18 X) == 1 == 100 X/(142+18 X)MOLES AND WE HAVE 56/18 MOLES OF H2O (AS 56% LOST GOES IN H20). EQUATING THEM YOU GET X = 10 MOLES ..... AND THUS CALCULTE MOL WT. YOURSELF.
Na2SO4.XH2O -> Na2SO4 + X H2O (BECOMES ANHYDROUS)
1MOLE == 1 == X MOLES (== MEANS REACT)
THUS,
100/(142+18 X) == 1 == 100 X/(142+18 X)MOLES
AND WE HAVE 56/18 MOLES OF H2O (AS 56% LOST GOES IN H20).
EQUATING THEM YOU GET X = 10 MOLES .....
AND THUS CALCULTE MOL WT. YOURSELF.
on heaing it looses all water molecules let n water molecules be present in the compound before heating its mol. wt. is : 18n + 142 loss in mol wt after heating : 18n 18n/(142+18n) = 54/100 on solving fo rn gives : n=10 the compound is NA2SO4.10H2O -- regards Ramesh
on heaing it looses all water molecules
let n water molecules be present in the compound before heating
its mol. wt. is : 18n + 142
loss in mol wt after heating : 18n
18n/(142+18n) = 54/100
on solving fo rn gives : n=10
the compound is NA2SO4.10H2O
--
regards
Ramesh
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