A) \[0.50\text{ }L\]of \[10\text{ }N\text{ }HCl\] and \[0.50\text{ }L\]of \[4\text{ }N\text{ }HCl\]
B) \[0.60\text{ }L\]of \[10\text{ }N\]\[HCl\] and \[0.40\text{ }L\]of \[4\text{ }N\text{ }HCl\]
C) \[0.80\text{ }L\]\[10\text{ }N\] \[HCl\]of and \[0.20\text{ }L\]of \[4\text{ }N\text{ }HCl\]
D) \[0.75\text{ }L\]of \[10\text{ }N\] \[HCl\] and \[0.25\text{ }L\]of \[4\text{ }N\text{ }HCl\]
Correct Answer: A
Solution :
Let V litre of \[10\text{ }N\]\[HCl\] be mixed with \[(1-V)\] litre of \[4\text{ }N\]\[HCl\] to give \[(V+1-V)=1\,L\] 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 \[10N\,\,HCl=0.50L\] Volume of \[4\text{ }N\text{ }HCl=1-0.50=0.50\text{ }L\]
You need to login to perform this action.
You will be redirected in
3 sec
