Question

If $100$ mL of $1$ M $H_{2}S{O}_4$ solution is mixed with $100$ ml of $98$% (w/w) of $H_{2}S{O}_4$ solution $(d=0.1gm{L}^{-1})$, then :

• A. concentration of solution becomes half.
• B. volume of solution becomes $200$ ml.
• C. mass of $H_{2}S{O}_4$ in the solution is $98$ g.
• D. mass of $H_{2}S{O}_4$ in the solution is $19.6$ g.

Answer 1

Answer: (B), (D)

The correct options are
B volume of solution becomes $200$ ml.
D mass of $H_{2}S{O}_4$ in the solution is $19.6$ g.

$100$ mL of $1$ M $H_{2}S{O}_4+100mL$ ($98$%, $d=0.1)H_{2}S{O}_4$
$=100\times 1M+100mL\times \left(\frac{98\times 10\times 0.1}{98}\right)$
$=100\times 1M+100\times 1M$
$\therefore \left[H_{2}S{O}_4\right]=\frac{100\times 1+100\times 1}{200mL}=1M=98g/1000mL$
Mass of $H_{2}S{O}_4=\frac{98g\times 200mL}{1000mL}=19.6g$
Concentration of each component becomes half of the initial value.
Hence, options B and D are correct.