A) 3.476 g
B) 12.38 g
C) 34.76 g
D) 1.238 g
Correct Answer: A
Solution :
\[KMn{{O}_{4}}\]= Mohr salt \[\frac{{{M}_{1}}{{V}_{1}}}{1}=\frac{{{M}_{2}}{{V}_{2}}}{5}=\left[ \frac{W}{M\times V}\times 1000 \right]\times \frac{{{V}_{2}}}{5}\] \[\left[ \frac{W\times 1000}{58\times 1000} \right]\times 18\] \[=\frac{3.92\times 1000}{392\times 1000}\times \frac{20}{5}\] W=3.476gm/L.You need to login to perform this action.
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