Question

$100ml$ of $5\text{ m}{\text{H}}_2{\text{SO}}_4$ of density $1gm/ml$ is mixed with $100ml$ of $8\text{ m}{\text{H}}_2{\text{SO}}_4$ of density $1.25g/mL$ . 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 $solution=200ml$
$\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}=\frac{\text{m}\text{o}\text{l}\text{a}\text{l}\text{i}\text{t}\text{y}\times \text{d}\times 1000}{\left(\text{m}\text{o}\text{l}\text{a}\text{l}\text{i}\text{t}\text{y}\times {\left(\text{M}\right)}^{{}^{\prime }}+1000\right)}$
or
$\frac{1}{\text{M}\text{o}\text{l}\text{a}\text{r}\text{i}\text{t}\text{y}}=\frac{1}{\text{d}}\left(\frac{1}{\text{m}\text{o}\text{l}\text{a}\text{l}\text{i}\text{t}\text{y}}+\frac{{\left(\text{M}\right)}^{{}^{\prime }}}{1000}\right)$
Let molarity of $5\text{ m}{\text{H}}_2{\text{SO}}_4={\text{M}}_1$
$\frac{1}{M_1}=\frac{1}{d}\left(\frac{1}{m}+\frac{M\prime }{1000}\right)$
Where $M\prime =mol$ 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 $8\text{m}{\text{H}}_2{\text{SO}}_4={\text{M}}_2$
$\frac{1}{M_2}=\frac{1}{1.25}\left(\frac{1}{8}+\frac{M\prime }{1000}\right)$
$=\frac{1}{1.25}\times \left(\frac{1}{8}+\frac{98}{1000}\right)$
or $\frac{1}{M_2}=\frac{1}{1.25}\times 0.223$
or ${\text{M}}_2=\frac{1.25}{0.223}=5.6\text{M}$
${\text{M}}_{\text{mix}\text{.}}=\frac{{\text{M}}_{\text{1}}{\text{V}}_1+{\text{M}}_{\text{2}}{\text{V}}_2}{{\text{V}}_{\text{1}}{\text{+V}}_{\text{2}}}$ $=4.5\text{ M}$