A 1800 gm mixture of anhydrous CuSO4 (s) and its hydrated form CuSO4⋅5H2O(s) undergoes 20% loss in mass on heating. Mole fraction of CuSO4 in mixture is (Atomic mass of Cu = 64):

nH2O=1800×0.218=20∴ nCuSO4⋅5H2O=4⇒ nCuSO4=800160=5XCuSO4=59

A 1800 gm mixture of anhydrous CuSO4 (s) and its hydrated form CuSO4⋅5H2O(s) undergoes 20% loss in mass on heating. Mole fraction of CuSO4 in mixture is (Atomic mass of Cu = 64):

nH2O=1800×0.218=20∴ nCuSO4⋅5H2O=4⇒ nCuSO4=800160=5XCuSO4=59

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A $1800 g m$ mixture of anhydrous ${CuSO}_{4}$ (s) and its hydrated form $[{CuSO}_{4} \cdot 5 H_{2} O (s)]$ undergoes $20 %$ loss in mass on heating. Mole fraction of ${CuSO}_{4}$ in mixture is (Atomic mass of $C u = 64$ ):

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$\frac{3}{40}$

$\frac{4}{9}$

$\frac{5}{9}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} n_{H_{2} O} = \frac{1800 \times 0.2}{18} = 20 \\ ∴ n_{{CuSO}_{4} \cdot 5 H_{2} O} = 4 \\ \Rightarrow n_{{CuSO}_{4}} = \frac{800}{160} = 5 \\ X {CuSO}_{4} = \frac{5}{9} \end{array}$

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