🤓 Based on our data, we think this question is relevant for Professor Li's class at CSUF.

$\overline{){\mathbf{molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{solution}}}$

mass percent of H_{2}SO_{4} = 40.0**%**

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{Mass}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

This means in **100 g of solution**, we have **40.0 g H _{2}SO_{4}**

$\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{\%}\mathbf{=}\frac{\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}{\mathbf{100}\mathbf{}\mathbf{g}\mathbf{}\mathbf{solution}}$

molar mass H_{2}SO_{4} = 98.079 g/mol

The moles of H_{2}SO_{4} is:

$\mathbf{moles}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}{\mathbf{98}\mathbf{.}\mathbf{079}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}$

**moles H _{2}SO_{4}**

Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H_{2}SO_{4}, for which the density is 1.3057 g/mL.

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