The volumes of $4\mathrm{N}\mathrm{HCl}$ and $10\mathrm{N}\mathrm{HCl}$ required to make $1L$ of $6\mathrm{N}\mathrm{HCl}$ are:

The correct option is C

$0.67L$of $4\mathrm{N}\mathrm{HCl}$and $0.33L$of $10\mathrm{N}\mathrm{HCl}$

The explanation of the correct option:

(C)$\mathbf{0}\mathbf{.}\mathbf{67}\mathbf{}\mathit{L}$ of $\mathbf{4}\mathbf{}\mathbf{N}\mathbf{}\mathbf{HCl}$ and $\mathbf{0}\mathbf{.}\mathbf{33}\mathbf{}\mathit{L}$ of $\mathbf{10}\mathbf{}\mathbf{N}\mathbf{}\mathbf{HCl}$

Step 1:

Given: Total volume ($V$) is $1L$ and normality ($N$) is $6N$

Step 2:

Let,

• $N_1$ be normality of $4\mathrm{N}\mathrm{HCl}$; $V_1$ is the volume of $4\mathrm{N}\mathrm{HCl}$
• $N_2$ be the normality of$10\mathrm{N}\mathrm{HCl}$; $V_2$ be the volume of $10\mathrm{N}\mathrm{HCl}$

Step 3:

• Now, $V_1+V_2=V$

$\therefore V_2=V-V_1$

• We know $N_1{V}_1+N_2{V}_2=NV$

Step 4:

• Substituting the values in the above formula we,

$$\begin{gathered} 4V_1+10(1-V_1)=6\times 1 \\ \therefore V_1=\frac{2}{3} \end{gathered}$$

• Similarly, $V_2=1-\frac{2}{3}=\frac{1}{3}$
• Thus, $V_1=0.67Land{V}_2=0.33L$

Explanation of the incorrect options:

Since, the correct volumes are $0.67L$of $4\mathrm{N}\mathrm{HCl}$and $0.33L$of $10\mathrm{N}\mathrm{HCl}$. Therefore, options (A), (B) and (D) stand incorrect.

Therefore, the correct option is (C)$\mathbf{0}\mathbf{.}\mathbf{67}\mathbf{}\mathit{L}$ of $\mathbf{4}\mathbf{}\mathbf{N}\mathbf{}\mathbf{HCl}$ and $\mathbf{0}\mathbf{.}\mathbf{33}\mathbf{}\mathit{L}$ of $\mathbf{10}\mathbf{}\mathbf{N}\mathbf{}\mathbf{HCl}$.