$5 \,ml$ of $N$ $HCl$, $20 \,ml$ of $N/2\,\,{H_2}S{O_4}$ and $30\, ml$ of $N/3\,\,HN{O_3}$ are mixed together and volume made to one litre. The normality of the resulting solution is
→You are given $500\, mL$ of $2\,N\, HCl$ and $500\, mL$ of $5\,N\, HCl$. What will be the maximum volume of $3\, M\, HCl$ that you can make from these two sotutions ? .............. $\mathrm{mL}$
→$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH - C{H_3}} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,}
\end{array}$ $\xrightarrow[dry\,ether]{Na}X\xrightarrow[\left( 2 \right)\,{{H}_{2}}{{O}_{2}},\overset{\Theta }{\mathop{O}}\,{{H}_{\left( aq \right)}}]{\left( 1 \right)\,B{{H}_{3}},THF}Y;\,Y$ is
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