Question

The mixture of $$CuS$$ (molar weight = $$\displaystyle M_{1}$$) and $$\displaystyle Cu_{2}S$$ (molecular weight =$$\displaystyle M_{2}$$) is oxidised by $$\displaystyle KMnO_{4}$$ (molecular weight = $$\displaystyle M_{4}$$) in acidic medium, the product obtained are $$\displaystyle Cu^{2+}$$ and $$SO_{2+}$$. The equivalent weights of $$\displaystyle CuS,Cu_{2}S$$ and $$\displaystyle KMnO_{4}$$respectively, are :

A
M16,M28,M35
B
M13,M28,M35
C
M16,M27,M35
D
M16,M28,M37

Solution

The correct option is A $$\displaystyle \frac{M_{1}}{6},\frac{M_{2}}{8},\frac{M_{3}}{5}$$For $$CuS$$, the oxidation number  of $$S$$ changes from -2 to +4.Total change in the oxidation number is +6.Equivalent weight is the ratio of molar mass to the change in the oxidation number.$$\therefore$$ The equivalent weight of $$CuS$$ is $$\displaystyle \dfrac {M_1}{6}$$ For $$\displaystyle Cu_2S$$ the oxidation number  of S changes from -2 to +4. The oxidation number of Cu changes from +1 to +2.Total change in the oxidation number is $$\displaystyle +6+2(+1)=+8$$ .$$\therefore$$ The equivalent weight of $$\displaystyle Cu_2S$$ is $$\displaystyle \dfrac {M_2}{8}$$ For $$\displaystyle KMnO_4$$ , the oxidation number of Mn changes from +7 to +2.The change in the oxidation number is +5.$$\therefore$$ The equivalent weight of $$\displaystyle KMnO_4$$  is $$\displaystyle \dfrac {M_3}{5}$$ Chemistry

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