Question

$480ml$ of $1.5M$ non-electrolyte solution is mixed with $520ml$ of $1.2M$ non-electrolyte solution. What is the molarity of the final mixture?

• A. $2.70M$
• B. $1.344M$
• C. $1.50M$
• D. $1.20M$

Answer 1

The correct option is B $1.344M$

Given,

$V_1=480mL$, $M_1=1.5M$

We know,

$\Rightarrow M_1=\frac{n_1}{V_1}$ [Molarity formula]

$\Rightarrow n_1=M_1{V}_1=1.5\times 480\times {10}^{-3}L$

$\Rightarrow n_1=720\times {10}^{-3}$ moles

Also, $V_2=520mL$, $M_2=1.2M$

We know $\Rightarrow M_2=\frac{n_2}{V_2}$

$\Rightarrow n_2=1.2\times 520\times {10}^{-3}$

$\Rightarrow n_2=624\times {10}^{-3}$ moles

$\therefore$ Total no. of moles in the solution $=1.344$

Also, total volume of the mixture $=480+520=1000mL=1L$

Now,

Molarity $=\frac{no.ofmoles}{1L}$

$=\frac{1.344}{1}M$

$\therefore$ Molarity of final mixture $=1.344M$

Hence, the correct option is $B$