A) 80ml.
B) 60 mL
C) 40 mL
D) 90 mL
Correct Answer: B
Solution :
Key Idea : (i) \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] (ii) Amount of water to be added = total volume - volume of \[NaOH\] Given normality of \[NaOH={{N}_{1}}=0.1\,N\] Volume of \[NaOH={{V}_{1}}=?\] Normality of \[HCl\,{{N}_{2}}=0.2\,N\] Volume of \[HCl\,{{V}_{2}}=50\,\,mL\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[0.1\times {{V}_{1}}=0.2\times 50\] \[{{V}_{1}}=\frac{0.2\times 50}{0.1}=100\,\,mL\] V of \[NaOH=40\,\,mL\] Amount \[{{H}_{2}}O\] to be added = 100 - 40 - 60 mL
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