Question

The volumes of a 10 N HCl solution and a 4 N HCl solution required to make 1 L of a 7 N HCl solution are:

• A. 0.50 L of 10 N HCl and 0.50 L of 4 N HCl
• B. 0.60 L of 10 N HCl and 0.40 L of 4 N HCl
• C. 0.80 L of 10 N HCl and 0.20 L of 4 N HCl
• D. 0.75 L of 10 N HCl and 0.25 L of 4 N HCl

Answer 1

The correct option is A 0.50 L of 10 N HCl and 0.50 L of 4 N HCl

$N_1{V}_1+N_2{V}_2=NV$
Where $N_1$ and $N_2$ are the normalities of the given HCl solutions and $V_1$ and $V_2$ are their respective volumes.
$N_1$ = 10 N, $N_2$ = 4 N
$N\Rightarrow$ The normality of the final solution = 7 N
$V\Rightarrow$ The volume of the final solution = 1 L
$V_1+V_2=1L\Rightarrow V_2=(1-V_1)L\Rightarrow 10V_1+4(1-V_1)=1\times 7$
$\Rightarrow 10V_1+4-4V_1=7\Rightarrow 6V_1=3$
Therefore, $V_1=\frac{3}{6}=0.5$ L
$V_2=1–0.5$ = 0.5 L