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Question: A \[1800g\] mixture of anhydrous \[CuS{O_4}\] and its hydrated for \[\left( {CuS{O_4}.5{H_2}O} \righ...

A 1800g1800g mixture of anhydrous CuSO4CuS{O_4} and its hydrated for (CuSO4.5H2O)\left( {CuS{O_4}.5{H_2}O} \right)undergoes 20%20\% loss in mass on heating. Mole fraction of CuSO4CuS{O_4} in mixture is (atomic mass of CuCu is 6464)
A.340\dfrac{3}{{40}}
B.49\dfrac{4}{9}
C.59\dfrac{5}{9}
D.12\dfrac{1}{2}

Explanation

Solution

We need to know that the anhydrous copper sulphate is a chemical compound having the formula CuSO4.5H2OCuS{O_4}.5{H_2}O which is white in colour. And if the water is added to the copper sulphate, it will hydrate and it is converted into blue in colour. Here, the copper ions trap the individual water molecules into the ionic lattice. A reaction between water and copper sulphate can be used for the test of water.

Complete answer:
The mole fraction of the mixture present in the water is not equal to 340\dfrac{3}{{40}}. Hence, option (A) is incorrect.
The mole fraction of copper sulphate which is present in the mixture after the heating of mixture is not equal to 49\dfrac{4}{9}. Hence, the option (B) is incorrect.
According to the question, the reaction of heating anhydrous copper sulphate can be written as,
CuSO4.5H2OCuSO4+5H2OCuS{O_4}.5{H_2}O \to CuS{O_4} + 5{H_2}O
Given, the amount of anhydrous copper sulphate mixture is equal to1800g1800g and loss of water in % is equal to 2020.
Therefore, loss of water in mass =20100×1800 = \dfrac{{20}}{{100}} \times 1800
=360g= 360g
Number of moles of water =36018=20moles = \dfrac{{360}}{{18}} = 20moles
Moles of copper sulphate eliminated from anhydrous form =205=4moles = \dfrac{{20}}{5} = 4moles
And the molar mass of copper sulphate =96+64=160g = 96 + 64 = 160g
Mass of copper sulphate =4×160=640g = 4 \times 160 = 640g
Mass of anhydrous copper sulphate initially=640+360=1000g = 640 + 360 = 1000g
Hence, mass of copper sulphate at initial time =18001000=800g = 1800 - 1000 = 800g
Thus, moles of copper sulphate =800160=5moles = \dfrac{{800}}{{160}} = 5moles
And the total moles of copper sulphate5+4=9moles5 + 4 = 9moles
Therefore, mole fraction of CuS{O_4}$$$$,{\chi _{CuS{O_4}}} = \dfrac{5}{{5 + 4}} = \dfrac{5}{9}
Hence, option (C) is correct.
The mole fraction of copper sulphate in the mixture will not equal 12\dfrac{1}{2}. Hence, the option (D) is incorrect.

Hence, option (C) is correct.

Note:
We need to know that the mole fraction is the unit of the amount of constituent which is denoted as, ni{n_i}and it is divided with the total amount of the constituents which is present in a mixture. When adding water into the copper sulphate, there is a formation of copper sulphate pentahydrate and the water molecule will be lost by the dehydration of copper sulphate and there is a formation of copper sulphate monohydrate.