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 \[NaOH=40g\] \[\therefore \] \[40\,g\]of\[NaOH=1\,g\]eq. of\[NaOH\] \[\therefore \] \[0.275g\]of \[NaOH\]\[=\frac{1}{40}\times 0.275\,eq.\] \[=\frac{1}{40}\times 0.275\times 1000\] \[=6.88\,meq\] \[\because \] \[\underset{\text{(HCl)}}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{\text{(NaOH)}}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[{{N}_{1}}\times 35.4=6.88\] \[(\because \,meq=NV)\] \[\therefore \] \[{{N}_{1}}=0.194\]
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