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 , M , HCl 2 , M , HCl Mixing ratio: 4:1 Let's denote the volume of 1 , M , HCl as 4V and the volume of 2 , M , HCl as V. The final volume of the mixture is: V_final = 4V + V = 5V The total moles of HCl in the mixed solution are: Moles from 1 , M , HCl = 1 x 4V = 4V Moles from 2 , M , HCl = 2 x V = 2V Total moles of HCl = 4V + 2V = 6V The final molarity (M_final) is given by the total moles of solute divided by the total volume of the solution: M_final = Total moles of HCl / V_final = 6V / 5V = 6 / 5 = 1.2 , M Thus, the final molarity of the HCl solution is: 1.2 So, the correct answer is: C. 1.2
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:

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

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

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 $2 \text{ M HCl}$: $$ 2 \text{ M } \times V = 2V \text{ moles} $$

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

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

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