Question

$100 \, ml$ of $5 \text{ m} \, \text{H}_{2} \text{SO}_{4}$ of density $1 \, g m / m l$ is mixed with $100 \, ml$ of $8 \text{ m} \, \text{H}_{2} \text{SO}_{4}$ of density $1.25 \, g / m L$ . If there is no change in volume of resulting solution due to mixing, the molarity of the resulting mixture is-

• A. $5.5 \, \text{M}$
• B. $6.5 \, \text{M}$
• C. $4.5 \text{ M}$
• D. $5.26 \, \text{M}$

Answer 1

Answer: (C)

On mixing total vol of $solution=200 \, ml$
$\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y} \, =\frac{\text{m} \text{o} \text{l} \text{a} \text{l} \text{i} \text{t} \text{y} \times \text{d} \times 1000}{\left(\text{m} \text{o} \text{l} \text{a} \text{l} \text{i} \text{t} \text{y} \, \times \left(\text{M}\right)^{'} + 1000\right)}$
or
$\frac{1}{\text{M} \text{o} \text{l} \text{a} \text{r} \text{i} \text{t} \text{y}}=\frac{1}{\text{d}}\left(\frac{1}{\text{m} \text{o} \text{l} \text{a} \text{l} \text{i} \text{t} \text{y}} + \frac{\left(\text{M}\right)^{'}}{1000}\right)$
Let molarity of $5 \text{ m} \, \text{H}_{2} \text{SO}_{4} = \text{M}_{1}$
$\frac{1}{M_{1}}=\frac{1}{d}\left(\frac{1}{m} + \frac{M ′}{1000}\right)$
Where $M′=mol$ wt. of $H_{2}S O_{4}$
$\frac{1}{M_{1}}=\frac{1}{1}\left(\frac{1}{5} + \frac{98}{1000}\right)=0.298\sim eq0.3$
$M_{1}=10 / 3$
Let molarity of $8 \, \text{m} \, \text{H}_{2} \text{SO}_{4} = \text{M}_{2}$
$\frac{1}{M_{2}}=\frac{1}{1.25}\left(\frac{1}{8} + \frac{M ′}{1000}\right)$
$=\frac{1}{1.25}\times \left(\frac{1}{8} + \frac{98}{1000}\right)$
or $\frac{1}{M_{2}}=\frac{1}{1.25} \, \times \, 0.223$
or $\text{M}_{2} = \frac{1.25}{0.223} = 5.6 \, \text{M}$
$\text{M}_{\text{mix} \text{.}}=\frac{\text{M}_{\text{1}} \text{V}_{1} + \text{M}_{\text{2}} \text{V}_{2}}{\text{V}_{\text{1}} \text{+V}_{\text{2}}}$ $= 4.5 \text{ M}$