QJEEM05Chd3d3: JEE Main 2005 Chemistry - Class 11 JEEM05Chd3d3 Solution: Two solutions of a substance (non electrolyte) are mixed in the …

Short Cut Method : Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.

Here molarity of first solution is 1.2 and molarity of second solution is 1.5.
So the molarity of the mixture will always be more than 1.2 and less than 1.5. Remember the molarity of mixture can't be either 1.2 or 1.5.

So from the option you an see only 1.344 M can be the right answer.

Normal method :
The formula for molarity of mixture of two substance is = ${{{M_1}{V_1} + {M_2}{V_2}} \over {{V_1} + {V_2}}}$

Here ${M_1} = 1.5$ , ${V_1} = 480$ , ${M_2} = 1.2$ , ${V_2} = 520$

$\therefore$ Molarity of mixture = ${{1.5 \times 480 + 1.2 \times 520} \over {480 + 520}}$

= ${{720 + 624} \over {1000}}$

= ${{1344} \over {1000}}$

= 1.344