Question

If 100 ml of $1M{H}_{2}S{O}_4$ solution is mixed with 100 ml of $9.8\%$ (w/w) $H_{2}S{O}_4$ solution ($d=1g/ml$) then:

• A. $\text{Molarity of resulting solution is 1}$
• B. $\text{Volume of solution becomes 200 ml}$
• C. $\text{Mass of}{H}_{2}S{O}_4\text{in the solution is 98 g}$
• D. $\text{Mass of}{H}_{2}S{O}_4\text{in the solution is 19.6 g}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\text{Molarity of resulting solution is 1}$
B $\text{Volume of solution becomes 200 ml}$
D $\text{Mass of}{H}_{2}S{O}_4\text{in the solution is 19.6 g}$

$w=98g$
$M=98g$
$d=1g/ml$
we know the formula of Molarity is
$M=\frac{numberofmolesofsolute}{Volumeofsolution\left(inliter\right)}$
and number of mole is given as
$n=\frac{massofsolute}{molarmassofsolute}$
(A)$M_1=1M{V}_1=\frac{100}{1000}=0.1L$
$9.8g$ of $H_{2}S{O}_4$ present in 100 g of solution.
Since, $d=1gm{l}^{-1}\therefore V_2=100ml=0.1L$
Sinec, $9.8g$ is in $100ml$ of solution, So, $98g$ will be present in $1000ml$ or $1L$ of solution.
Molarity $M_2=\frac{\text{Moles of solute}}{\text{Volume}}{M}_2=\frac{98}{98\times 1}=1M$

$M_3=\frac{M_1{V}_1+M_2{V}_2}{V_1+V_2}{M}_3=\frac{(1\times 0.1)+(1\times 0.1)}{0.2}=1M$

(B) Volume of resulting solution$=100+100=200$ ml
(D) Using molarity formula,
$\text{Mass of}{H}_{2}S{O}_4=\frac{200\times 1}{1000}\times 98=19.6$ g.