A) 0.0267
B) 0.2670
C) 1.0267
D) 1.1670
Correct Answer: B
Solution :
Normality of a mixture \[(N)=\frac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\] Normality \[({{N}_{1}})\] of \[{{H}_{2}}S{{O}_{4}}=\]molarity \[\times \] basicity \[=0.2\times 2=0.4\,N\] \[{{N}_{2}}=0.2\times 1=0.2\,N\] \[{{V}_{1}}=100\,mL,{{V}_{2}}=200\,mL\] \[N=\frac{0.4\times 100+0.2\times 200}{100+200}\] \[=\frac{40+40}{300}=\frac{80}{300}=0.2670\,N\] Normality of a mixture of acid and base (N') \[N'=\frac{{{N}_{1}}{{V}_{1}}\sim {{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\]You need to login to perform this action.
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