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16.1 g of a hydrated metal sulphate was heated to constant mass. After cooling to room temperature, the residual
anhydrous metal sulphate weighted 7.1 g. How many moles of water of crystallization are there in one mole of
the hydrated metal sulphate? (Relative atomic mass : anhydrous metal sulphate = 142.0 ; water = 18.0)
A.  4   B.  5   C.  7   D.  10
Ans solution plz

2.60 g of a metal X combine with 1.20 g of oxygen to form an oxide in which the oxidation number
of X is +3. What is the relative atomic mass of X? (Relative atomic mass : O = 16.0)
A. 11.6 B. 34.7 C. 52.0 D. 104
Ans:C WHY?

sunnytsang

First question:
Let there be n moles of water.
16.1/(18n+142)=7.1/142.0 (By considering number of moles of metal sulphate)
So 18n+142=322, and n=10.
Therefore the answer is D.


Second question:
The oxide formed is X2O3. The ratio of X to O2 is 4:3.
Let the relative atomic mass of X be y.
3(2.60/y)=4(1.20/32.0) (By considering the mole ratio of X and O2 is 4:3)
So 7.80/y=0.15, and y=52.
Therefore the answer is C.