A) 1.5 N
B) 0.15 N
C) 0.066 N
D) 0.66 N
Correct Answer: C
Solution :
Volume of monobasic acid \[=10\text{ }c{{m}^{3}}\] Normality of monobasic acid = 0.1 N Volume of \[NaOH\]solution \[=15\text{ }c{{m}^{3}}\] Normality of \[NaOH\]solution = ? \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] (for monobasic acid) (for \[NaOH\]) \[10\times 0.1\,N=15\times {{N}_{2}}\] \[{{N}_{2}}=\frac{1\,N}{15}=0.066\,\,N\]
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