Question

To find the final molarity of the mixed HCl solution, we will use the concept of mixing solutions with different molarities in a given volume ratio.

We have:

  • $1 , \text{M} , \text{HCl}$
  • $2 , \text{M} , \text{HCl}$
  • Mixing ratio: $4:1$

Let's denote the volume of $1 , \text{M} , \text{HCl}$ as $4V$ and the volume of $2 , \text{M} , \text{HCl}$ as $V$.

The final volume of the mixture is: $$ V_{\text{final}} = 4V + V = 5V $$

The total moles of HCl in the mixed solution are: $$ \text{Moles from } 1 , \text{M} , \text{HCl} = 1 \times 4V = 4V $$ $$ \text{Moles from } 2 , \text{M} , \text{HCl} = 2 \times V = 2V $$ $$ \text{Total moles of HCl} = 4V + 2V = 6V $$

The final molarity ($M_{\text{final}}$) is given by the total moles of solute divided by the total volume of the solution: $$ M_{\text{final}} = \frac{\text{Total moles of HCl}}{V_{\text{final}}} = \frac{6V}{5V} = \frac{6}{5} = 1.2 , \text{M} $$

Thus, the final molarity of the HCl solution is: $$ \boxed{1.2} $$

So, the correct answer is:

  • C. 1.2

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Answer

:

To determine the final molarity of the solution obtained by mixing $1 \text{ M HCl}$ and $2 \text{ M HCl}$ in a $4:1$ volume ratio, we can follow these steps:

  1. Let the volume of $1 \text{ M HCl}$ be $4V$.

  2. Let the volume of $2 \text{ M HCl}$ be $V$.

  3. The total volume of the mixed solution is $5V$.

    The moles of HCl in $4V$ volume of $1 \text{ M HCl}$: $$ 1 \text{ M } \times 4V = 4V \text{ moles} $$

    The moles of HCl in $V$ volume of $2 \text{ M HCl}$: $$ 2 \text{ M } \times V = 2V \text{ moles} $$

  4. Total moles of $\text{HCl}$ in the final solution: $$ 4V \text{ moles} + 2V \text{ moles} = 6V \text{ moles} $$

  5. Total volume of the final solution: $$ 5V $$

  6. Molarity of the final HCl solution: $$ \frac{6V \text{ moles}}{5V} = 1.2 \text{ M} $$

Answer: C (1.2 M)


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