A) \[0.50L\]of \[10N\text{ }HCl\]and \[0.50L\]of \[4N\text{ }HCl\]
B) \[0.60L\] of \[10N\text{ }HCl\] and \[0.40L\]of \[4N\text{ }HCl\]
C) \[0.80L\]of \[10N\text{ }HCl\]and \[0.20L\]of \[4N\text{ }HCl\]
D) \[0.75L\]of \[10N\text{ }HCl\] and \[0.25L\]of \[4N\text{ }HCl\]
Correct Answer: A
Solution :
Let V litre of \[10N\text{ }HCl\]be mixed with \[(1-V)\] litre of \[4N\text{ }HCl\]to give \[(V+1-V)=1L\] of \[7N\,HCl\]. \[{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}=NV\] \[10V+4(1-V)=7\times 1\] \[10V+4-4V=7\] \[6V=7-4\] \[V=\frac{3}{6}=0.50L\] Volume of \[10N\text{ }HCl=0.50\text{ }L\] Volume of \[4N\text{ }HCl=1-0.50=0.50L\]
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