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  4. What volume of 6 M HCL should be added to 2 M HCL to get 1 L of 3 M HCL ?

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Q. What volume of $6\, M$ HCL should be added to $2\, M$ HCL to get $1\, L$ of $3\, M HCL$ ?

Some Basic Concepts of Chemistry

Solution:

Suppose the volume of $6\, M\, HCL$ required to obtain $1\, L$ of $3\, M$
$HCl =x L$
$\therefore $ volume of $2\, N\, HCl$ required $=(1-x) L$
Applying the molarity equation
$M_{1} V_{1}+M_{2} V_{2}=M_{3} V_{3}$
$6\, M\, HCl +2\, MHCl \,3\, M HCl$
$6 x+2(1-x)=3 \times 1$
$4 x=1$
$x=0.25\, L$
Hence, volume of $6\,M\, HCl$ required $=0.25\, L$
and volume of $2\,M\, HCl$ required $= 0.75\, L$