Question

Concentrated sulfuric acid (18.3 M) has a density of 1.84 g/mL. (a) How many moles of H₂SO₄ are in each milliliter of solution? (b) What is the mass % of H₂SO₄ in the solution?

Short answer

1. The mass of H₂SO₄ are in each milliliter of solution is$1.83\times {10}^{-2}\mathrm{mol}{\mathrm{H}}_2{\mathrm{SO}}_4/\mathrm{mL}\mathrm{solution}.$
2. The mass % of H₂SO₄ in the solution is 97.5 % of H₂SO₄.

Step-by-step solution

Finding the mass of H2SO4 are in each milliliter of solution

On multiplying the given molarity,

$\begin{array}{l}{\text{Mass\hspace{0.17em}of\hspace{0.17em}HNO}}_{\text{3}}\text{\hspace{0.17em}(1\hspace{0.17em}L\hspace{0.17em}solution)\hspace{0.33em}=\hspace{0.33em}}\frac{{\text{18.3\hspace{0.17em}mol\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1\hspace{0.17em}L\hspace{0.17em}solution}}\text{×}\frac{{\text{10}}^{\text{-3}}\text{L\hspace{0.17em}soln}}{\text{1\hspace{0.17em}mL\hspace{0.17em}solution}}\\ {\text{Mass\hspace{0.17em}of\hspace{0.17em}HNO}}_{\text{3}}{\text{\hspace{0.17em}(1\hspace{0.17em}L\hspace{0.17em}solution)\hspace{0.33em}=\hspace{0.33em}1.83×10}}^{\text{-2}}{\text{\hspace{0.17em}mol\hspace{0.33em}\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{/mL\hspace{0.17em}soln}\end{array}$

Finding the mass % of H2SO4 in the solution

On calculating the mass of H₂SO₄,

$\begin{array}{l}{\text{Mass\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{{\text{98.08\hspace{0.17em}g\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1\hspace{0.17em}mol\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{{\text{18.3\hspace{0.17em}mol\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1\hspace{0.17em}L\hspace{0.17em}soln}}\text{×}\frac{{\text{10}}^{\text{-3}}\text{L\hspace{0.17em}sln}}{\text{1\hspace{0.17em}mL\hspace{0.17em}sln}}\text{×}\frac{\text{1\hspace{0.17em}mL\hspace{0.17em}soln}}{\text{1.84\hspace{0.17em}g\hspace{0.17em}soln}}\text{×100\hspace{0.17em}g\hspace{0.17em}soln}\\ {\text{Mass\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=97.7\hspace{0.17em}g}\end{array}$