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