A) 80 mL
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 \[NaOH={{N}_{1}}=0.1\text{ }N\] Volume of \[NaOH={{V}_{1}}=?\] Normality of\[~HCl\text{ }{{N}_{2}}-0.2\text{ }N\] 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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