A) 0.50 L of\[10\text{ }N\text{ }HCl\]and 0.50 L of\[4\text{ }N\text{ }HCl\]
B) 0.60 L of\[10\text{ }N\text{ }HCl\]and 0.40 L of\[4\text{ }N\text{ }HCl\]
C) 0.80 L of\[10\text{ }N\text{ }HCl\]and 0.20 L of\[4\text{ }N\text{ }HCl\]
D) 0.75 L of\[10\text{ }N\text{ }HCl\]and 0.25 L of\[4\text{ }N\text{ }HCl\]
Correct Answer: A
Solution :
Let V litre of\[10\text{ }N\text{ }HCl\]be mixed with\[(1-V)\]litre of\[4\text{ }N\text{ }HCl\]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 \[10\text{ }N\text{ }HCl=0.50\text{ }L\] 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
