Question

$V_1$ mL of $a$M $HCl$ and $V_2$ mL of $b$ M $HCl$ are mixed to produce $M_{1}HCl$. Again $V_2$ mL of aM $HCl$ and $V_1$ mL of M $HCl$ are mixed to produce $M_{2}HCl$.

If $\frac{V_1}{V_2}$> 1 and $\frac{V_1}{V_2}=\frac{a}{b}$ and $\frac{M_1}{M_2}=\frac{5}{4}$, then the ratio $\frac{V_1}{V_2}$ is equal to:

• A. $2:1$
• B. $4:1$
• C. $1:2$
• D. $3:1$

Answer 1

The correct option is A $2:1$

$M_i{V}_i+M_j{V}_j=M_k(V_i+V_j)$

here

$M_i=aM$

$M_j=bM$

case 1:

$a{V}_1+b{V}_2=M_1(V_1+V_2)$ ----- eqn(1)

case 2:

$a{V}_2+b{V}_1=M_2(V_1+V_2)$ ----- eqn(2)

Given, $\frac{V_1}{V_2}=\frac{a}{b}$ -----eqn(3)

divide equation 1 and 2 by $V_2$ and substitute eqn(3)

we get

$eqn\left(1\right)÷V_2\Rightarrow a\frac{V_1}{V_2}+b=M_1(\frac{V_1}{V_2}+1)$

$\Rightarrow \frac{a^2}{b}+b=M_1(\frac{a}{b}+1)$ ---eqn (4)

$eqn\left(2\right)÷V_2\Rightarrow =a+b\frac{V_1}{V_2}=M_2(\frac{V_1}{V_2}+1)$

$\Rightarrow 2a=M_2(\frac{a}{b}+1)$ ------eqn(5)

$eqn\left(4\right)÷eqn\left(5\right)\Rightarrow \frac{a^2+b^2}{2ab}=\frac{M_1}{M_2}=\frac{5}{4}$

$\Rightarrow 4(\frac{a}{b}{)}^2-10(\frac{a}{b})+4=0$

$\Rightarrow \frac{a}{b}=\frac{10\pm \sqrt{100-64}}{8}=\frac{10\pm 6}{8}$

$\Rightarrow \frac{a}{b}=2:1$ or $1:2$

But its given that $\frac{V_1}{V_2}>1$

Hence $\frac{a}{b}=2:1$