A) \[60g\]
B) \[6g\]
C) \[4g\]
D) \[40g\]
Correct Answer: B
Solution :
\[1500\text{ }c{{m}^{3}}\]of \[0.1\text{ }N\,HCl\]have number of gm equivalence \[=\frac{{{N}_{1}}\times {{V}_{1}}}{1000}=\frac{1500\times 0.1}{1000}=0.15\] \[\because \] \[0.15\text{ }gm\]. equivalent of \[NaOH\] \[=1.5\times 40=6gm\].
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