CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image