The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be

We are required to find the molarity of the final mixture formed by combining two hydrochloric acid solutions of different concentrations. The fundamental idea is that when two solutions are mixed, the total number of moles of solute present in the mixture equals the sum of the moles originally present in each solution. After mixing, the new molarity is obtained by dividing this total number of moles by the total volume of the mixture.

First, recall the definition of molarity (M):

$$\text{Molarity (M)} \;=\; \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}.$$

Hence, the number of moles in any given solution is given by the product of its molarity and its volume in litres. We shall apply this one‐step formula separately to both HCl solutions, then add the results.

For the first solution, we have 750 mL of 0.5 M HCl.

Convert the volume from millilitres to litres:

$$750 \text{ mL} \;=\; \frac{750}{1000} \text{ L} \;=\; 0.750 \text{ L}.$$

Now calculate the moles of HCl present in this portion by using $$n = M \times V$$:

$$n_1 \;=\; 0.5 \text{ M} \times 0.750 \text{ L}.$$

Perform the multiplication step by step:

$$0.5 \times 0.750 \;=\; 0.375.$$

So,

$$n_1 = 0.375 \text{ mol}.$$

For the second solution, we have 250 mL of 2 M HCl.

Once again, convert the given volume to litres:

$$250 \text{ mL} \;=\; \frac{250}{1000} \text{ L} \;=\; 0.250 \text{ L}.$$

Using $$n = M \times V$$, determine the moles in this second portion:

$$n_2 \;=\; 2 \text{ M} \times 0.250 \text{ L}.$$

Compute the multiplication:

$$2 \times 0.250 \;=\; 0.500.$$

Thus,

$$n_2 = 0.500 \text{ mol}.$$

Now add the moles from the two individual solutions to obtain the total number of moles in the mixture:

$$n_{\text{total}} \;=\; n_1 + n_2 \;=\; 0.375 \text{ mol} + 0.500 \text{ mol}.$$

Carrying out the addition:

$$n_{\text{total}} = 0.875 \text{ mol}.$$

Next, find the total volume after mixing by adding the separate volumes (expressed in litres):

$$V_{\text{total}} \;=\; 0.750 \text{ L} + 0.250 \text{ L} = 1.000 \text{ L}.$$

Finally, determine the new molarity $$M_{\text{mix}}$$ of the combined solution using the definition of molarity:

$$M_{\text{mix}} \;=\; \frac{n_{\text{total}}}{V_{\text{total}}}.$$

Substituting the numerical values we just calculated:

$$M_{\text{mix}} \;=\; \frac{0.875 \text{ mol}}{1.000 \text{ L}}.$$

Dividing by 1, we get directly:

$$M_{\text{mix}} = 0.875 \text{ M}.$$

This value matches option C in the provided list.

Hence, the correct answer is Option C.