A) \[0.97\,\,N\]
B) \[0.142\,\,N\]
C) \[0.194\,\,N\]
D) \[0.244\,\,N\]
Correct Answer: C
Solution :
We know that \[1\,\,g\] equivalent weight of \[NaOH=40\,\,g\] \[\therefore \] \[40\,\,g\]of\[NaOH=1\,\,g\] eq. of\[NaOH\] \[\therefore \] \[0.275\,\,g\]of\[NaOH=\frac{1}{40}\times 0.275\,\,eq.\] \[=\frac{1}{40}\times 0.275\times 1000\] \[=6.88\,\,meq\] \[\underset{(HCl)}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{(NaOH)}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[{{N}_{1}}\times 35.4=6.88\] \[(\because \,\,meq=NV)\] \[{{N}_{1}}=0.194\]
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