Question

$100\text{mL}$ of an $H_{2}S{O}_4$ solution that has a molarity of $1\text{M}$ and density of $1.50\text{g/mL}$ is mixed with $400\text{mL}$ of water. If the final density of the resulting solution is $1.25\text{g/mL}$, the final molarity of the $H_{2}S{O}_4$ solution is:

• A. $4.40\text{M}$
• B. $0.145\text{M}$
• C. $0.520\text{M}$
• D. $0.227\text{M}$

Answer 1

The correct option is D $0.227\text{M}$

No. of moles of $H_{2}S{O}_4$ reacted = molarity x volume
= 1 M x 100mL
= 0.1 moles
Weight of $H_{2}S{O}_4$ = density x volume
=1.50 x 100 = 150 g
Weight of water reacted with $H_{2}S{O}_4=400g$
Total mass of the resulting solution$H_{2}S{O}_4=150+400g=550g$
Total volume of the resulting solution$H_{2}S{O}_4=\frac{550}{1.25}=440mL$
Molarity =$\frac{no.ofmoles}{volume}=\frac{0.1}{440}\times 1000=0.227M$