Question

100 mL of an $\mathrm{H}_2\mathrm{SO}_4$ solution that has a molarity of $1 , \mathrm{M}$ and density of $1.50 , \mathrm{g/mL}$ is mixed with 400 mL of water. If the final density of the resulting solution is $1.25 , \mathrm{g/mL}$, the final molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution is:

A) $4.40 , \mathrm{M}$
B) $0.145 , \mathrm{M}$
C) $0.520 , \mathrm{M}$
D) $0.227 , \mathrm{M}$

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Answer

The correct option is D) $0.227 , \mathrm{M}$

To determine the final molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution, follow these steps:

  1. Calculate the number of moles of $\mathrm{H}_2\mathrm{SO}_4$ initially present:

    Given:

    • Molarity ($M$) of solution: $1 , \mathrm{M}$
    • Volume ($V$) of solution: $100 , \mathrm{mL} = 0.1 , \mathrm{L}$

    Number of moles = Molarity $\times$ Volume $$ 1 , \mathrm{M} \times 0.1 , \mathrm{L} = 0.1 , \mathrm{moles} $$

  2. Calculate the weight of $\mathrm{H}_2\mathrm{SO}_4$:

    Given:

    • Density ($d$) of initial solution: $1.50 , \mathrm{g/mL}$
    • Volume ($V$) of solution: $100 , \mathrm{mL}$

    Weight = Density $\times$ Volume $$ 1.50 , \mathrm{g/mL} \times 100 , \mathrm{mL} = 150 , \mathrm{g} $$

  3. Determine the weight of water added:

    Given:

    • Volume of water: $400 , \mathrm{mL}$
    • Density of water: $1.00 , \mathrm{g/mL}$ (approximately)

    Weight of added water: $$ 400 , \mathrm{mL} \times 1 , \mathrm{g/mL} = 400 , \mathrm{g} $$

  4. Calculate the total mass of the resulting solution:

    Total mass = Mass of $\mathrm{H}_2\mathrm{SO}_4$ solution + Mass of added water $$ 150 , \mathrm{g} + 400 , \mathrm{g} = 550 , \mathrm{g} $$

  5. Determine the total volume of the resulting solution:

    Given:

    • Final density of the solution: $1.25 , \mathrm{g/mL}$

    Total volume = Total mass / Final density $$ \frac{550 , \mathrm{g}}{1.25 , \mathrm{g/mL}} = 440 , \mathrm{mL} $$

  6. Calculate the final molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution:

    Molarity = $\text{Number of moles} / \text{Total volume}$ $$ \text{Molarity} = \frac{0.1 , \mathrm{moles}}{440 , \mathrm{mL}} \times 1000 , \mathrm{mL/L} $$ $$ = \frac{0.1 , \mathrm{moles}}{0.44 , \mathrm{L}} $$ $$ = 0.227 , \mathrm{M} $$

Thus, the final molarity of the $\mathrm{H}_2\mathrm{SO}_4$ solution is $0.227 , \mathrm{M}$.


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