A) 10 mL
B) 20 mL
C) 5 mL
D) 15 mL
Correct Answer: C
Solution :
The volume of \[\text{0}\text{.1 M}\,\text{Ca(OH}{{\text{)}}_{\text{2}}}\] required to neutralise 10 mL of \[\text{0}\text{.1 N HCl}\] \[\underset{Ca{{(OH)}_{2}}}{\mathop{2{{M}_{1}}{{V}_{1}}}}\,=\underset{(HCl)}{\mathop{{{M}_{2}}{{V}_{2}}}}\,\] (because \[\text{0}\text{.1}\,\text{N}\,\text{HCl}\,\text{=}\,\text{0}\text{.1}\,\text{M}\,\text{HCl}\]) \[2\times 0.1\times {{V}_{1}}=0.1\times 10\] \[{{V}_{1}}=\frac{0.1\times 10}{2\times 0.1}\] \[\therefore \] \[{{V}_{1}}=5\,mL\]
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