Two solutions of a substance (non-electrolyte) are mixed in the following manner 480 mℓ of 1.5 M of first solution with 520 mℓ of 1.2 M of second solution. The molarity of final solution is :

Question

Two solutions of a substance (non-electrolyte) are mixed in the following manner 480 mℓ of 1.5 M of first solution with 520 mℓ of 1.2 M of second solution. The molarity of final solution is : (A) 1.20 M (B) 1.50 M (C) 1.344 M (D) 2.70 M

Tardigrade Admin · Accepted Answer

For I mM = 480 × 1.5 = 720 For II mM = 520 × 1.2 = 624 Total milli-mole = 720 + 624 = 134 ∴ Molarity = (1344/480 + 520) = 1.344 M

Q. Two solutions of a substance (non-electrolyte) are mixed in the following manner $480 m ℓ$ of $1.5$ $M$ of first solution with $520 m ℓ$ of $1.2$ $M$ of second solution. The molarity of final solution is :

$1.20$ $M$

$1.50$ $M$

$1.344$ $M$

$2.70$ $M$

For $I$ $m M$ $= 480 \times 1.5 = 720$
For $II$ $m M = 520 \times 1.2 = 624$
Total milli-mole $= 720 + 624 = 134$
$∴$ Molarity $= \frac{1344}{480 + 520} = 1.344 M$

Q. Two solutions of a substance (non-electrolyte) are mixed in the following manner 480mℓ of 1.5 M of first solution with 520mℓ of 1.2 M of second solution. The molarity of final solution is :

1.20 M

1.50 M

1.344 M

2.70 M