Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?

Some Basic Concepts of Chemistry PYQ — JEE Main 2005 · Shift 0 · Q53 with Solution

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Correct answer: B

For mixing solutions of the same non-electrolyte, total moles are added:

$n_{\text{total}} = n_1 + n_2 = M_1V_1 + M_2V_2$

and final molarity is

$M_f = \frac{n_{\text{total}}}{V_{\text{total}}}$

$V_1 = 480\,\text{mL} = 0.480\,\text{L}$ $V_2 = 520\,\text{mL} = 0.520\,\text{L}$

For the first solution:

$n_1 = M_1V_1 = 1.5 \times 0.480 = 0.72\,\text{mol}$

For the second solution:

$n_2 = M_2V_2 = 1.2 \times 0.520 = 0.624\,\text{mol}$

$n_{\text{total}} = 0.72 + 0.624 = 1.344\,\text{mol}$

$V_{\text{total}} = 0.480 + 0.520 = 1.000\,\text{L}$

$M_f = \frac{1.344}{1.000} = 1.344\,\text{M}$

$\boxed{1.344\,\text{M}}$

So the correct option is B.