A) 10 mL
B) 20 mL
C) 5 mL
D) 15 mL
Correct Answer: C
Solution :
The volume of\[0.1\text{ }M\text{ }Ca{{(OH)}_{2}}\]required to neutralise 10 mL of\[0.1\text{ }N\text{ }HCl\] \[\underset{Ca{{(OH)}_{2}}}{\mathop{2{{M}_{1}}{{V}_{1}}}}\,=\underset{(HCl)}{\mathop{{{M}_{2}}{{V}_{2}}}}\,\] (because\[0.1\text{ }N\text{ }HCl=0.1\text{ }M\text{ }HCl\]) \[2\times 0.1\times {{V}_{1}}=0.1\times 10\] \[{{V}_{1}}=\frac{0.1\times 10}{2\times 0.1}\] \[\therefore \] \[{{V}_{1}}=5\,mL\]
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