Question

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

• A. 0.25 L
• B. 1.00 L
• C. 0.75
• D. 2.50 L

Answer 1

Answer: (A)

Suppose the volume of $6MHCL$ required to obtain $1L$ of $3M$
$HCl=xL$
$\therefore$ volume of $2NHCl$ required $=(1-x)L$
Applying the molarity equation
$M_1{V}_1+M_2{V}_2=M_3{V}_3$
$6MHCl+2MHCl3MHCl$
$6x+2(1-x)=3\times 1$
$4x=1$
$x=0.25L$
Hence, volume of $6MHCl$ required $=0.25L$
and volume of $2MHCl$ required $=0.75L$