Question

$250\text{mL}$ of a $‘x^{\prime }\text{M}$ solution and $500\text{mL}$ of a $‘y^{\prime }\text{M}$ solution of a solute $A$ are mixed and diluted to $2$ litres to produce a solution with the final concentration $1.6\text{M}$. If $x:y=5:4$, then the value of $x$ and $y$ respectively are:

• A. $4.92$ and $4.92$
• B. $3.94$ and $3.152$
• C. $4.92$ and $3.94$
• D. $4.92$ and $3.936$

Answer 1

The correct option is C $4.92$ and $3.94$

Initial conditions of the solution:
Solute A:
(i) Volume of solution $V_1$ = $250\text{mL}$
Molarity of solution $M_1$ = $x\text{M}$

(ii) Volume of solution $V_2$ = $500\text{mL}$
Molarity of solution $M_2$ = $y\text{M}$

Final conditions of solution :
Molarity of solution $M_f$ = $1.6\text{M}$
Volume of solution $V_f$ = $2\text{L}$
Also,
$\frac{x}{y}=\frac{5}{4}$
$\Rightarrow x=\frac{5}{4}y...\left(i\right)$ .

We know that, in case of dilution, moles remain conserved.
$M_1{V}_1+M_2{V}_2=M_f{V}_f$
$\Rightarrow x\frac{250}{1000}+y\frac{500}{1000}=1.6\times 2$
$\Rightarrow (250\times x)+(500\times y)=3200$
On simplifying ,
$x+2y=12.8...\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$ we get
$1.25y+2y=12.8$
$\Rightarrow 3.25y=12.8$
$\Rightarrow y=3.94$ and $x=\frac{5}{4}y$
$\therefore x=1.25\times 3.94=4.92$
Hence, the molarity of the $250\text{mL}$ solution is $4.92\text{M}$ and the molarity of the $500\text{mL}$ solution is $3.94\text{M}$