Mass of hydrate + crucible = 47.29 g
Mass of anhydrous salt = 2.7 g
Molar mass of anhydrous salt CuSO4 = 159.5 g
Given,
mass of empty crucible = 42.45 g
mass of hydrate salt= 4.84 g
mass of crucible after first heating = 46.1 g
mass of crucible after second heating= 45.153 g
mass of crucible after third heating= 45.15 g
so, as per the question we need to find...
Mass of hydrate + crucible = ? g
Mass of anhydrous salt = ? g
Molar mass of anhydrous salt CuSO4 = ? g
∴Mass of hydrate + crucible = 42.45 + 4.48 = 47.29 g
The given salt is in hydrate form, to remove water from this molecule we need to perform heating .
So we are taking the substance into the crucible as it is in less quantity.
Here, we performed heating 3 times and note the weight after every heating.
After this, assume that the water is totally evaporated and the remaining salt is in anhydrous form,
∴ Mass of anhydrous salt = 45.15 - 42.45 = 2.7 g
To find the molar mass of anhydrous salt of CuSO4,
atomic weight of Cu = 63.5 g
atomic weight of S = 32 g
atomic weight of O =16 g
∴ molar mass of anhydrous salt of CuSO4 = 63.5 + 32 + (16 ×3)
=159.5 gm
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