A) 0.1 N
B) 0.2 N
C) 0.3 N
D) 0.4 N
Correct Answer: B
Solution :
Normality of the resulting mixture \[=\frac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}+{{N}_{3}}{{V}_{3}}}{{{V}_{1}}+{{V}_{2}}+{{V}_{3}}}\] \[=\frac{75\times \frac{1}{5}+10\times \frac{1}{2}+30\times \frac{1}{10}}{75+10+30}\] \[=\frac{15+5+3}{115}\] \[=\frac{23}{115}=\frac{N}{5}=0.2N\]You need to login to perform this action.
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