A) 70 ml
B) 32 ml
C) 35 ml
D) 16 ml
Correct Answer: D
Solution :
Volume m of HCl neutralised by NaOH = (Caustic soda) = \[{{V}_{1}}\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]; \[0.1\times {{V}_{1}}=0.2\times 30\]; \[{{V}_{1}}=60ml\] V total (HCl) = 100ml \[{{V}_{1}}\] = 60ml _________________ 40ml 40ml 0.1N HCl is now neutralised by KOH (0.25N) \[\to \] (HCl) \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\](KOH) \[0.1\times 40=0.25\times {{V}_{2}}\]; \[{{V}_{2}}=16ml\].You need to login to perform this action.
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