When 1.615g anhydrous salt was placed in moist air it's mass was found to be 2.875g.The percentage amount of elements in anhydrous salt is as follows. Zn=40.6, S=19.8 and O=39.6. If complete hydration of salt is taking place , then calculate the no. of molecules of water of crystallization
Answers
Given:
The mass of anhydrous salt = 1.615 gm
The mass of salt after hydration = 2.875 gm
% Zn = 40.6 %
% S = 19.8 %
% O = 39.6 %
To Find:
The no of molecules of water of crystallization in the salt.
Calculation:
- 100 gm of given anhydrous salt will contain 40.6 gm Zn, 19.8 gm S and 39.6 gm O.
- No of moles of Zn = 40.6/65 = 0.625
- No of moles of S = 19.8/32 = 0.618
- No of moles of O = 39.6/16 = 2.475
⇒ Zn : S : O = 1 : 1 : 4
- So, the formula of the anhydrous salt is ZnSO4.
- The molar mass of ZnSO4 = 161 gm
- The no of moles of ZnSO4 in 1.615 gm = 1.615/161 = 0.01
- The mass of water = 2.875 - 1.615 = 1.26 gm
- The no of moles of water in 1.26 gm = 1.26/18 = 0.07
⇒ The no of moles of water absorbed by 0.01 moles of anhydrous salt = 0.07
⇒The no of moles of water absorbed by 1 mole of anhydrous salt = 0.07/0.01 = 7