The problem involves mixing two HCl solutions of different molarities in a given volume ratio. Here's the stepbystep solution: Given: 1 M HCl 2 M HCl Volume ratio of 4:1 Let the volume of 1 M HCl be 4V and the volume of 2 M HCl be 1V. The total volume V_total of the mixture is: V_total = 4V + 1V = 5V The number of moles of HCl from each solution: From 1 M HCl: Moles = Molarity x Volume = 1 x 4V = 4V From 2 M HCl: Moles = Molarity x Volume = 2 x 1V = 2V The total number of moles of HCl in the mixture: Total moles = 4V + 2V = 6V The molarity of the final solution is: Molarity = Total moles / Total volume = 6V / 5V = 1.2 M Thus, the final molarity of the HCl solution is: C. 1.2
Question
The problem involves mixing two $\mathrm{HCl}$ solutions of different molarities in a given volume ratio. Here's the stepbystep solution:
Given:
 $1 \mathrm{M}$ $\mathrm{HCl}$
 $2 \mathrm{M}$ $\mathrm{HCl}$
 Volume ratio of $4:1$
Let the volume of $1 \mathrm{M}$ $\mathrm{HCl}$ be $4V$ and the volume of $2 \mathrm{M}$ $\mathrm{HCl}$ be $1V$.
The total volume $V_{\text{total}}$ of the mixture is: $$ V_{\text{total}} = 4V + 1V = 5V $$
The number of moles of $\mathrm{HCl}$ from each solution:
 From $1 \mathrm{M}$ $\mathrm{HCl}$: $$ \text{Moles} = \text{Molarity} \times \text{Volume} = 1 \times 4V = 4V $$
 From $2 \mathrm{M}$ $\mathrm{HCl}$: $$ \text{Moles} = \text{Molarity} \times \text{Volume} = 2 \times 1V = 2V $$
The total number of moles of $\mathrm{HCl}$ in the mixture: $$ \text{Total moles} = 4V + 2V = 6V $$
The molarity of the final solution is: $$ \text{Molarity} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{6V}{5V} = 1.2 \mathrm{M} $$
Thus, the final molarity of the $\mathrm{HCl}$ solution is:
C. $1.2$
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Answer
To determine the molarity of the resulting $\mathrm{HCl}$ solution when mixing $1 \mathrm{M}$ $\mathrm{HCl}$ and $2 \mathrm{M}$ $\mathrm{HCl}$ in a $4:1$ volume ratio:

Calculate the moles of $\mathrm{HCl}$ in each solution being mixed:

For $1 \mathrm{M}$ $\mathrm{HCl}$:
Let the volume be $4V$, thus total moles = $1 \mathrm{M} \times 4V = 4V$.

For $2 \mathrm{M}$ $\mathrm{HCl}$:
Let the volume be $V$, thus total moles = $2 \mathrm{M} \times V = 2V$.


Add the moles together to find the total moles of $\mathrm{HCl}$ in the mixture: $$ \text{Total moles} = 4V + 2V = 6V $$

Determine the total volume of the mixed solution: $$ \text{Total volume} = 4V + V = 5V $$

Calculate the final molarity of the solution: $$ [\mathrm{HCl}] = \frac{\text{Total moles}}{\text{Total volume}} = \frac{6V}{5V} = 1.2 \mathrm{M} $$
Therefore, the final molarity of the $\mathrm{HCl}$ solution is $1.2 \mathrm{M}$.
Final Answer: C
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